RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 5 - Trigonometric Functions

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RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 5 Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Trigonometric Functions are extremely popular among class 12 Commerce students for Maths Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020 2021 Volume 1 Book of class 12 Commerce Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020 2021 Volume 1 Solutions. All RD Sharma XI 2020 2021 Volume 1 Solutions for class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 5.18:

Question 1:

stion Prove the following identities (1-16)
$\sec^{4} θ - \sec^{2} θ = \tan^{4} θ + \tan^{2} θ$

Answer:

$LHS = \sec^{4} x - \sec^{2} x = \sec^{2} x (\sec^{2} x - 1) = (\tan^{2} x + 1) (\tan^{2} x) (∵ \sec^{2} x - \tan^{2} x = 1) = \tan^{4} x + \tan^{2} x = RHS Hence proved .$

Page No 5.18:

Question 2:

Prove the following identities (1-16)
$\sin^{6} x + \cos^{6} x = 1 - 3 \sin^{2} x \cos^{2} x$

Answer:

$LHS = \sin^{6} x + \cos^{6} x = {(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3} = (\sin^{2} x + \cos^{2} x) [{(\sin^{2} x)}^{2} + {(\cos^{2} x)}^{2} - \sin^{2} x \cos^{2} x] [∵ a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - ab)] = 1 \times [{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x - \sin^{2} x \cos^{2} x] [∵ \sin^{2} x + \cos^{2} x = 1 and a^{2} + b^{2} = {(a + b)}^{2} - 2 ab] = 1^{2} - 3 \sin^{2} x \cos^{2} x = 1 - 3 \sin^{2} x \cos^{2} x = RHS Hence proved .$

Page No 5.18:

Question 3:

Prove the following identities (1-16)

$(cosec x - \sin x) (\sec x - \cos x) (\tan x + \cot x) = 1$

Answer:

$LHS = (cosec x - \sin x) (\sec x - \cos x) (\tan x + \cot x) = (\frac{1}{\sin x} - \sin x) (\frac{1}{\cos x} - \cos x) (\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}) = (\frac{1 - \sin^{2} x}{\sin x}) (\frac{1 - \cos^{2} x}{\cos x}) (\frac{\sin^{2} x + \cos^{2} x}{\cos x \sin x}) = (\frac{\cos^{2} x}{\sin x}) (\frac{\sin^{2} x}{\cos x}) (\frac{1}{\cos x \sin x}) = 1 = RHS Hence proved .$

Page No 5.18:

Question 4:

Prove the following identities (1-16)
$cosec x (\sec x - 1) - \cot x (1 - \cos x) = \tan x - \sin x$

Answer:

$LHS = cosec x (\sec x - 1) - \cot x (1 - \cos x) = \frac{1}{\sin x} (\frac{1}{\cos x} - 1) - \frac{\cos x}{\sin x} (1 - \cos x) = \frac{1}{\sin x} (\frac{1 - \cos x}{\cos x}) - \frac{\cos x}{\sin x} (1 - \cos x) = (\frac{1 - \cos x}{\sin x}) (\frac{1}{\cos x} - \cos x) = (\frac{1 - \cos x}{\sin x}) (\frac{1 - \cos^{2} x}{\cos x}) = (\frac{1 - \cos x}{\sin x}) (\frac{\sin^{2} x}{\cos x}) = (1 - \cos x) (\frac{\sin x}{\cos x}) = \frac{\sin x}{\cos x} - \sin x = \tan x - \sin x = RHS Hence proved .$

Page No 5.18:

Question 5:

Prove the following identities (1-16)
$\frac{1 - \sin x \cos x}{\cos x (\sec x - cosec x)} \cdot \frac{\sin^{2} x - \cos^{2} x}{\sin^{3} x + \cos^{3} x} = \sin x$

Answer:

$LHS = \frac{1 - \sin x \cos x}{\cos x (\sec x - cosec x)} \times \frac{\sin^{2} x - \cos^{2} x}{\sin^{3} x + \cos^{3} x} = \frac{1 - \sin x \cos x}{\cos x (\frac{1}{\cos x} - \frac{1}{\sin x})} \times \frac{{(\sin x)}^{2} - {(\cos x)}^{2}}{{(\sin x)}^{3} + {(\cos x)}^{3}} = \frac{1 - \sin x \cos x}{\cos x (\frac{\sin x - \cos x}{\cos x \sin x})} \times \frac{(\sin x + \cos x) (\sin x - \cos x)}{(\sin x + \cos x) [{(\sin x)}^{2} + {(\cos x)}^{2} - \sin x \cos x]} = \frac{\sin x (1 - \sin x \cos x)}{(\sin x - \cos x)} \times \frac{(\sin x + \cos x) (\sin x - \cos x)}{(\sin x + \cos x) [{(\sin x)}^{2} + {(\cos x)}^{2} - \sin x \cos x]} = \frac{\sin x (1 - \sin x \cos x)}{1} \times \frac{1 \times 1}{[\sin^{2} x + \cos^{2} x - \sin x \cos x]} = \sin x (1 - \sin x \cos x) \times \frac{1}{(1 - \sin x \cos x)} = \sin x = RHS Hence proved .$

Page No 5.18:

Question 6:

Prove the following identities (1-16)

$\frac{\tan x}{1 - \cot x} + \frac{\cot x}{1 - \tan x} = (\sec x cossec x + 1)$

Answer:

$LHS = \frac{\tan x}{1 - \cot x} + \frac{\cot x}{1 - \tan x} = \frac{\frac{\sin x}{\cos x}}{1 - \frac{\cos x}{\sin x}} + \frac{\frac{\cos x}{\sin x}}{1 - \frac{\sin x}{\cos x}} = \frac{\frac{\sin x}{\cos x}}{\frac{\sin x - \cos x}{\sin x}} + \frac{\frac{\cos x}{\sin x}}{\frac{\cos x - \sin x}{\cos x}} = \frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x - \cos x} + \frac{\cos x}{\sin x} \times \frac{\cos x}{\cos x - \sin x} = \frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x - \cos x} + \frac{\cos x}{\sin x} \times \frac{\cos x}{- (\sin x - \cos x)} = \frac{\sin^{2} x}{\cos x (\sin x - \cos x)} - \frac{\cos^{2} x}{\sin x (\sin x - \cos x)} = \frac{\sin^{3} x - \cos^{3} x}{\sin x \cos x (\sin x - \cos x)} = \frac{(\sin x - \cos x) [\sin^{2} x + \cos^{2} x + \sin x \cos x]}{\sin x \cos x (\sin x - \cos x)} = \frac{1 \times [1 + \sin x \cos x]}{\sin x \cos x} = \frac{1 + \sin x \cos x}{\sin x \cos x} = \frac{1}{\sin x \cos x} + \frac{\sin x \cos x}{\sin x \cos x} = \frac{1}{\sin x} \times \frac{1}{\cos x} + 1 = cosec x \times \sec x + 1 = (\sec x co \sec x + 1) = RHS Hence proved .$

Page No 5.18:

Question 7:

Prove the following identities (1-16)
$\frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x} + \frac{\sin^{3} x - \cos^{3} x}{\sin x - \cos x} = 2$

Answer:

$LHS = \frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x} + \frac{\sin^{3} x - \cos^{3} x}{\sin x - \cos x} = \frac{(\sin x + \cos x) [\sin^{2} x + \cos^{2} x - \sin x \cos x]}{(\sin x + \cos x)} + \frac{(\sin x - \cos x) [\sin^{2} x + \cos^{2} x + \sin x \cos x]}{(\sin x - \cos x)} = [\sin^{2} x + \cos^{2} x - \sin x \cos x] + [\sin^{2} x + \cos^{2} x + \sin x \cos x] = [1 - \sin x \cos x] + [1 + \sin x \cos x] = 2 = RHS Hence proved .$

Page No 5.18:

Question 8:

Prove the following identities (1-16)
${(\sec x \sec y + \tan x \tan y)}^{2} - {(\sec x \tan y + \tan x \sec y)}^{2} = 1$

Answer:

$LHS = {(\sec x \sec y + \tan x \tan y)}^{2} - {(\sec x \tan y + \tan x \sec y)}^{2} = [{(\sec x \sec y)}^{2} + {(\tan x \tan y)}^{2} - 2 (\sec x \sec y) (\tan x \tan y)] - [{(\sec x \tan y)}^{2} + {(\tan x \sec y)}^{2} - 2 (\sec x \tan y) (\tan x \sec y)] = [\sec^{2} x \sec^{2} y + \tan^{2} x \tan^{2} y - 2 \sec x \sec y \tan x \tan y] - [\sec^{2} x \tan^{2} y + \tan^{2} x \sec^{2} y - 2 \sec x \sec y \tan x \tan y] = \sec^{2} x \sec^{2} y + \tan^{2} x \tan^{2} y - 2 \sec x \sec y \tan x \tan y - \sec^{2} x \tan^{2} y - \tan^{2} x \sec^{2} y + 2 \sec x \sec y \tan x \tan y = \sec^{2} x \sec^{2} y - \sec^{2} x \tan^{2} y + \tan^{2} x \tan^{2} y - \tan^{2} x \sec^{2} y = \sec^{2} x (\sec^{2} y - \tan^{2} y) + \tan^{2} x (\tan^{2} y - \sec^{2} y) = \sec^{2} x (\sec^{2} y - \tan^{2} y) - \tan^{2} x (\sec^{2} y - \tan^{2} y) = \sec^{2} x \times 1 - \tan^{2} x \times 1 = \sec^{2} x - \tan^{2} x = 1 = RHS Hence proved .$

Page No 5.18:

Question 9:

Prove the following identities (1-16)
$\frac{\cos x}{1 - \sin x} = \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}$

Answer:

$RHS = \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} = \frac{(1 + \cos x) + (\sin x)}{(1 + cosx) - (\sin x)} = \frac{[(1 + \cos x) + (\sin x)] [(1 + \cos x) + (\sin x)]}{[(1 + \cos x) - (\sin x)] [(1 + \cos x) + (\sin x)]} = \frac{{[(1 + \cos x) + (\sin x)]}^{2}}{{(1 + \cos x)}^{2} - {(\sin x)}^{2}} = \frac{{(1 + \cos x)}^{2} + {(\sin x)}^{2} + 2 (1 + \cos x) (\sin x)}{1^{2} + \cos^{2} x + 2 \times 1 \times \cos x - \sin^{2} x} = \frac{1 + \cos^{2} x + 2 \cos x + \sin^{2} x + 2 \sin x \cos x + 2 \sin x}{1 + \cos^{2} x + 2 \cos x - \sin^{2} x} = \frac{1 + (\sin^{2} x + \cos^{2} x) + 2 \cos x + 2 \sin x \cos x + 2 \sin x}{(1 - \sin^{2} x) + \cos^{2} x + 2 \cos x} = \frac{1 + 1 + 2 \cos x + 2 \sin x \cos x + 2 \sin x}{\cos^{2} x + \cos^{2} x + 2 \cos x} = \frac{2 + 2 \cos x + 2 \sin x \cos x + 2 \sin x}{2 \cos^{2} x + 2 \cos x} = \frac{1 + \cos x + \sin x \cos x + \sin x}{\cos^{2} x + \cos x} = \frac{1 (1 + \cos x) + \sin x (\cos x + 1)}{\cos x (\cos x + 1)} = \frac{(\cos x + 1) (1 + \sin x)}{\cos x (cosx + 1)} = \frac{(1 + \sin x)}{\cos x} = \frac{(1 + \sin x) \times \cos x}{\cos x \times \cos x} = \frac{(1 + \sin x) \times \cos x}{\cos^{2} x} = \frac{(1 + \sin x) \times \cos x}{1 - \sin^{2} x} = \frac{(1 + \sin x) \times \cos x}{(1 + \sin x) (1 - \sin x)} = \frac{\cos x}{1 - \sin x} = LHS Hence proved .$

Page No 5.18:

Question 10:

Prove the following identities (1-16)

$\frac{\tan^{3} x}{1 + \tan^{2} x} + \frac{\cot^{3} x}{1 + \cot^{2} x} = \frac{1 - 2 \sin^{2} x \cos^{2} x}{\sin x \cos x}$

Answer:

$LHS = \frac{\tan^{3} x}{1 + \tan^{2} x} + \frac{\cot^{3} x}{1 + \cot^{2} x} = \frac{\tan^{3} x}{\sec^{2} x} + \frac{\cot^{3} x}{{cosec}^{2} x} = \frac{\frac{\sin^{3} x}{\cos^{3} x}}{\frac{1}{\cos^{2} x}} + \frac{\frac{\cos^{3} x}{\sin^{3} x}}{\frac{1}{\sin^{2} x}} = \frac{\sin^{3} x}{\cos^{3} x} \times \frac{\cos^{2} x}{1} + \frac{\cos^{3} x}{\sin^{3} x} \times \frac{\sin^{2} x}{1} = \frac{\sin^{3} x}{\cos x} + \frac{\cos^{3} x}{\sin x} = \frac{\sin^{4} x + \cos^{4} x}{\sin x \cos x} = \frac{{(\sin^{2} x)}^{2} + {(\cos^{2} x)}^{2}}{\sin x \cos x} = \frac{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x}{\sin x \cos x} = \frac{1^{2} - 2 \sin^{2} x \cos^{2} x}{\sin x \cos x} = \frac{1 - 2 \sin^{2} x \cos^{2} x}{\sin x \cos x} = RHS Hence proved .$

Page No 5.18:

Question 11:

Prove the following identities (1-16)

$1 - \frac{\sin^{2} x}{1 + \cot x} - \frac{\cos^{2} x}{1 + \tan x} = \sin x \cos x$

Answer:

$1 - \frac{\sin^{2} x}{1 + \cot x} - \frac{\cos^{2} x}{1 + \tan x} = \sin x \cos x$

$LHS = 1 - \frac{\sin^{3} x}{\sin x + \cos x} - \frac{\cos^{3} x}{\sin x + \cos x} = \frac{\sin x + \cos x - (\sin^{3} x + \cos^{3} x)}{\sin x + \cos x} = \frac{(\sin x + \cos x) (1 - \sin^{2} x - \cos^{2} x + \sin x \cos x)}{\sin x + \cos x} = (1 - \sin^{2} x - \cos^{2} x + \sin x \cos x) = (1 - 1 + \sin x \cos x) = \sin x \cos x$

= RHS
Hence proved.

Page No 5.18:

Question 12:

Prove the following identities (1-16)
$(\frac{1}{\sec^{2} x - \cos^{2} x} + \frac{1}{{cosec}^{2} x - \sin^{2} x}) \sin^{2} x \cos^{2} x = \frac{1 - \sin^{2} x \cos^{2} x}{2 + \sin^{2} x \cos^{2} x}$

Answer:

$(\frac{1}{\sec^{2} x - \cos^{2} x} + \frac{1}{{cosec}^{2} x - \sin^{2} x}) \sin^{2} x \cos^{2} x = \frac{1 - \sin^{2} x \cos^{2} x}{2 + \sin^{2} x \cos^{2} x}$

$\begin{array}{rcl} LHS & = & (\frac{1}{\sec^{2} x - \cos^{2} x} + \frac{1}{{cosec}^{2} x - \sin^{2} x}) \sin^{2} x \cos^{2} x \\ = & (\frac{1}{\frac{1}{\cos^{2} x} - \cos^{2} x} + \frac{1}{\frac{1}{\sin^{2} x} - \sin^{2} x}) \sin^{2} x \cos^{2} x \\ = & (\frac{\cos^{2} x}{1 - \cos^{4} x} + \frac{\sin^{2} x}{1 - \sin^{4} x}) \sin^{2} x \cos^{2} x \\ = & (\frac{\cos^{2} x (1 - \sin^{4} x) + \sin^{2} x (1 - \cos^{4} x)}{(1 - \cos^{4} x) (1 - \sin^{4} x)}) \sin^{2} x \cos^{2} x \\ = & (\frac{1 - \cos^{2} x \sin^{4} x - \cos^{4} x \sin^{2} x}{(1 + \sin^{2} x) (1 - s {in}^{2} x) (1 + \cos^{2} x) (1 - \cos^{2} x)}) \sin^{2} x \cos^{2} x \\ = & (\frac{1 - \cos^{2} x \sin^{4} x - \cos^{4} x \sin^{2} x}{(1 + \sin^{2} x) . \cos^{2} x . (1 + \cos^{2} x) . \sin^{2} x}) \sin^{2} x \cos^{2} x \\ = & (\frac{1 - \cos^{2} x \sin^{4} x - \cos^{4} x \sin^{2} x}{(1 + \sin^{2} x) (1 + \cos^{2} x)}) \\ = & \frac{1 - \cos^{2} x \sin^{2} x (\sin^{2} x + \cos^{2} x)}{2 + \sin^{2} x . \cos^{2} x} \\ = & \frac{1 - \cos^{2} x \sin^{2} x}{2 + \sin^{2} x . \cos^{2} x} \end{array}$

= RHS

Hence proved.

Page No 5.18:

Question 13:

Prove the following identities (1-17)

${(1 + \tan α \tan β)}^{2} + {(\tan α - \tan β)}^{2} = \sec^{2} α \sec^{2} β$

Answer:

${(1 + \tan α \tan β)}^{2} + {(\tan α - \tan β)}^{2} = \sec^{2} α \sec^{2} β$

$LHS = {(1 + \tan α \tan β)}^{2} + {(\tan α - \tan β)}^{2} = 1 + \tan^{2} α \tan^{2} β + 2 \tan α \tan β + \tan^{2} α + \tan^{2} β - 2 \tan α \tan β = 1 + \tan^{2} α \tan^{2} β + \tan^{2} α + \tan^{2} β = \tan^{2} α (\tan^{2} β + 1) + 1 (1 + \tan^{2} β) = (1 + \tan^{2} β) (1 + \tan^{2} α) = \sec^{2} α . \sec^{2} β = RHS$

Hence proved.

Page No 5.18:

Question 14:

Prove the following identities (1-16)

$\frac{(1 + \cot x + \tan x) (\sin x - \cos x)}{\sec^{3} x - {cosec}^{3} x} = \sin^{2} x \cos^{2} x$

Answer:

$\frac{(1 + \cot x + \tan x) (\sin x - \cos x)}{\sec^{3} x - {cosec}^{3} x} = \sin^{2} x \cos^{2} x$

$LHS = \frac{(1 + \cot x + \tan x) (\sin x - \cos x)}{\sec^{3} x - {cosec}^{3} x} = \frac{(1 + \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}) (\sin x - \cos x)}{\frac{1}{\cos^{3} x} - \frac{1}{\sin^{3} x}} = \frac{(\sin x \cos x + \cos^{2} x + \sin^{2} x) (\sin x - \cos x) (\sin^{2} x \cos^{2} x)}{(\sin^{3} x - \cos^{3} x)} = \frac{(1 + \sin x \cos x) (\sin x - \cos x) (\sin^{2} x \cos^{2} x)}{(\sin x - \cos x) (\sin^{2} x + \cos^{2} x + \sin x \cos x)} = \sin^{2} x \cos^{2} x = RHS Hence proved .$

Page No 5.18:

Question 15:

Prove the following identities (1-16)

$\frac{2 \sin x \cos x - \cos x}{1 - \sin x + \sin^{2} x - \cos^{2} x} = \cot x$

Answer:

$LHS = \frac{2 \sin x \cos x - \cos x}{1 - \sin x + \sin^{2} x - \cos^{2} x} = \frac{\cos x (2 \sin x - 1)}{2 \sin^{2} x - \sin x} (∵ 1 - \cos^{2} x = \sin^{2} x) = \frac{\cos x (2 \sin x - 1)}{\sin x (2 \sin x - 1)} = \cot x = RHS Hence proved .$

Page No 5.18:

Question 16:

Prove the following identities (1-16)

$\cos x (\tan x + 2) (2 \tan x + 1) = 2 \sec x + 5 \sin x$

Answer:

$LHS = \cos x (\tan x + 2) (2 \tan x + 1) = \cos x (2 \tan^{2} x + 5 \tan x + 2) = \cos x (\frac{2 \sin^{2} x}{\cos^{2} x} + \frac{5 \sin x}{\cos x} + 2) = \frac{2 \sin^{2} x + 5 \sin x \cos x + 2 \cos^{2} x}{\cos x} = \frac{2 + 5 \sin x \cos x}{\cos x} = 2 \sec x + 5 \sin x = RHS Hence proved .$

Page No 5.18:

Question 17:

If $x = \frac{2 \sin x}{1 + \cos x + \sin x}$ , then prove that $\frac{1 - \cos x + \sin x}{1 + \sin x}$ is also equal to a.

Answer:

Disclaimer: There is some error in the given question.
The question should have been
Question: If $a = \frac{2 \sin x}{1 + \cos x + \sin x}$ , then prove that $\frac{1 - \cos x + \sin x}{1 + \sin x}$ is also equal to a.
So, the solution is done accordingly.

Solution:
$a = \frac{2 \sin x}{1 + \sin x + \cos x} Rationalising the denominator : \frac{2 \sin x}{1 + \sin x + \cos x} \times \frac{(1 + \sin x) - \cos x}{(1 + \sin x) - \cos x} = \frac{2 \sin x \{(1 + \sin x) - \cos x\}}{{(1 + \sin x)}^{2} - \cos^{2} x} = \frac{2 \sin x \{(1 + \sin x) - \cos x\}}{1 + \sin^{2} x + 2 \sin x - \cos^{2} x} = \frac{2 \sin x \{(1 + \sin x) - \cos x\}}{2 \sin^{2} x + 2 \sin x} = \frac{2 \sin x \{(1 + \sin x) - \cos x\}}{2 \sin x (1 + \sin x)} = \frac{(1 + \sin x) - \cos x}{1 + \sin x} ∴ a = \frac{(1 + \sin x) - \cos x}{1 + \sin x}$

Hence proved.

Page No 5.18:

Question 18:

If $\sin x = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ , then the values of tan x, sec x and cosec x

Answer:

$\sin x = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} We know : \sin^{2} x + \cos^{2} x = 1 \cos^{2} x = 1 - \sin^{2} x = 1 - {(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})}^{2} = \frac{(a^{4} + b^{4} + 2 a^{2} b^{2}) - (a^{4} + b^{4} - 2 a^{2} b^{2})}{{(a^{2} + b^{2})}^{2}} = \frac{4 a^{2} b^{2}}{{(a^{2} + b^{2})}^{2}} \Rightarrow \cos x = \frac{2 a b}{(a^{2} + b^{2})}$

$\tan x = \frac{\sin x}{\cos x} = \frac{\frac{a^{2} - b^{2}}{a^{2} + b^{2}}}{\frac{2 a b}{a^{2} + b^{2}}} = \frac{a^{2} - b^{2}}{2 a b} \sec x = \frac{1}{\cos x} = \frac{a^{2} + b^{2}}{2 a b} cosec x = \frac{1}{\sin x} = \frac{a^{2} + b^{2}}{a^{2} - b^{2}}$

Page No 5.18:

Question 19:

If $\tan x = \frac{b}{a}$ , then find the values of $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}$ .

Answer:

$\tan x = \frac{b}{a} Now, \sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}} = \sqrt{\frac{1 + \tan x}{1 - \tan x}} + \sqrt{\frac{1 - \tan x}{1 + \tan x}} = \frac{\tan x + 1 + 1 - \tan x}{\sqrt{1 - \tan^{2} x}} = \frac{2}{\sqrt{1 - \tan^{2} x}} = \frac{2 \cos x}{\sqrt{\cos^{2} x - \sin^{2} x}}$

Page No 5.18:

Question 20:

If $\tan x = \frac{a}{b},$ show that $\frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ .

Answer:

$LHS : \frac{a \sin x - b \cos x}{a \sin x + b \cos x} Dividing by b \cos x : = \frac{\frac{a \tan x}{b} - 1}{\frac{a \tan x}{b} + 1} Substituting the value of \tan x = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = RHS$
Hence proved.

Page No 5.19:

Question 21:

If $cosec x - \sin x = a^{3}, \sec x - \cos x = b^{3}$ , then prove that $a^{2} b^{2} (a^{2} + b^{2}) = 1$ .

Answer:

$cosec x - \sin x = a^{3} ∴ \frac{1}{\sin x} - \sin = a^{3} \Rightarrow \frac{1 - \sin^{2} x}{\sin x} = a^{3} \Rightarrow \frac{\cos^{2} x}{\sin x} = a^{3} \Rightarrow a = {(\frac{\cos^{2} x}{\sin x})}^{\frac{1}{3}} . . . . (i) Also, \sec x - \cos x = b^{3} \Rightarrow \frac{1}{\cos x} - \cos = b^{3} \Rightarrow \frac{1 - \cos^{2} x}{\cos x} = b^{3} \Rightarrow \frac{\sin^{2} x}{\cos x} = b^{3} \Rightarrow b = {(\frac{\sin^{2} x}{\cos x})}^{\frac{1}{3}} . . . . . (ii) Now, LHS = a^{2} b^{2} (a^{2} + b^{2}) = {(a b)}^{2} (a^{2} + b^{2}) = {[{(\frac{\cos^{2} x}{\sin x})}^{\frac{1}{3}} {(\frac{\sin^{2} x}{\cos x})}^{\frac{1}{3}}]}^{2} [{({(\frac{\cos^{2} x}{\sin x})}^{\frac{1}{3}})}^{2} + {({(\frac{\sin^{2} x}{\cos x})}^{\frac{1}{3}})}^{2}] = {(\sin x \cos x)}^{\frac{2}{3}} [\frac{{(\cos^{2} x)}^{\frac{2}{3}}}{{(\sin x)}^{\frac{2}{3}}} + \frac{{(\sin^{2} x)}^{\frac{2}{3}}}{{(\cos x)}^{\frac{2}{3}}}] = {(\sin x \cos x)}^{\frac{2}{3}} [\frac{{(\cos^{3} x)}^{\frac{2}{3}} + {(\sin^{3} x)}^{\frac{2}{3}}}{{(\sin x)}^{\frac{2}{3}} {(\cos x)}^{\frac{2}{3}}}] = {(\sin x \cos x)}^{\frac{2}{3}} [\frac{\cos^{2} x + \sin^{2} x}{{(\sin x \cos x)}^{\frac{2}{3}}}] = 1 = RHS$

Page No 5.19:

Question 22:

If $\cot x (1 + \sin x) = 4 m and \cot x (1 - \sin x) = 4 n,$ prove that ${(m^{2} + n^{2})}^{2} = m n$ .

Answer:

$Given : 4 m = c o t x (1 + \sin x) a n d 4 n = c o t x (1 - \sin x) Multiplying both the equations : \Rightarrow 16 m n = c o t^{2} x (1 - \sin^{2} x) \Rightarrow 16 m n = c o t^{2} x . \cos^{2} x \Rightarrow m n = \frac{\cos^{4} x}{16 \sin^{2} x} (1) Squaring the given equation : 16 m^{2} = c o t^{2} x {(1 + \sin x)}^{2} a n d 16 n^{2} = c o t^{2} x {(1 - \sin x)}^{2} \Rightarrow 16 m^{2} - 16 n^{2} = c o t^{2} x (4 \sin x) \Rightarrow m^{2} - n^{2} = \frac{c o t^{2} x . \sin x}{4} Squaring both sides, {(m^{2} - n^{2})}^{2} = \frac{c o t^{4} x . \sin^{2} x}{16} \Rightarrow {(m^{2} - n^{2})}^{2} = \frac{\cos^{4} x}{16 \sin^{2} x} (2) From (1) and (2) : {(m^{2} - n^{2})}^{2} = m n Hence proved .$

Page No 5.19:

Question 23:

If $\sin x + \cos x = m$ , then prove that $\sin^{6} x + \cos^{6} x = \frac{4 - 3 {(m^{2} - 1)}^{2}}{4}$ , where $m^{2} \leq 2$

Answer:

$\sin x + \cos x = m (Given) To prove : \sin^{6} x + \cos^{6} x = \frac{4 - 3 (m^{2} - 1)^{2}}{4}, where m^{2} \leq 2 Proof : LHS : \sin^{6} x + \cos^{6} x = {(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3} = {(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x) = 1 - 3 \sin^{2} x \cos^{2} x RHS : \frac{4 - 3 (m^{2} - 1)^{2}}{4} = \frac{4 - 3 {[{(\sin x + \cos x)}^{2} - 1]}^{2}}{4} = \frac{4 - 3 {[\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1]}^{2}}{4} = \frac{4 - 3 {[\sin^{2} x - (1 - \cos^{2} x) + 2 \sin x \cos x]}^{2}}{4} = \frac{4 - 3 \times 4 \sin^{2} x \cos^{2} x}{4} = 1 - 3 \sin^{2} x \cos^{2} x LHS = RHS Hence proved$

Page No 5.19:

Question 24:

If $a = \sec x - \tan x and b = cosec x + \cot x$ , then shown that $a b + a - b + 1 = 0$ .

Answer:

$a = \sec x - \tan x And, b = cosec x + \cot x = \frac{1 - \sin x}{\cos x} And, b = \frac{1 + \cos x}{\sin x} Now, we have : a b + a - b + 1 (\frac{1 - \sin x}{\cos x}) (\frac{1 + \cos x}{\sin x}) + \frac{1 - \sin x}{\cos x} - (\frac{1 + \cos x}{\sin x}) + 1 = \frac{1 - \sin x + \cos x - \sin x \cos x + \sin x - \sin^{2} x - \cos x - \cos^{2} x + \sin x \cos x}{\sin x \cos x} = \frac{1 - \sin^{2} x - \cos^{2} x}{\sin x \cos x} = 0$

Hence proved.

Page No 5.19:

Question 25:

Prove the: $|\sqrt{\frac{1 - \sin x}{1 + \sin x}}| + |\sqrt{\frac{1 + \sin x}{1 - \sin x}}| = - \frac{2}{\cos x}, where \frac{π}{2} < x < π$

Answer:

$LHS = |\sqrt{\frac{1 - \sin x}{1 + \sin x}}| + |\sqrt{\frac{1 + \sin x}{1 - \sin x}}| = |\sqrt{\frac{(1 - \sin x) (1 - \sin x)}{(1 + \sin x) (1 - \sin x)}}| + |\sqrt{\frac{(1 + \sin x) (1 + \sin x)}{(1 - \sin x) (1 + \sin x)}}| = |\sqrt{\frac{(1 - \sin x) (1 - \sin x)}{(1 + \sin x) (1 - \sin x)}}| + |\sqrt{\frac{(1 + \sin x) (1 + \sin x)}{(1 - \sin x) (1 + \sin x)}}| = |\sqrt{\frac{{(1 - \sin x)}^{2}}{1 - \sin^{2} x}}| + |\sqrt{\frac{{(1 + \sin x)}^{2}}{1 - \sin^{2} x}}| = |\sqrt{\frac{{(1 - \sin x)}^{2}}{\cos^{2} x}}| + |\sqrt{\frac{{(1 + \sin x)}^{2}}{\cos^{2} x}}| = |\frac{1 - \sin x}{\cos x}| + |\frac{1 + \sin x}{\cos x}| = |\frac{1 - \sin x + 1 + \sin x}{\cos x}| = |\frac{2}{\cos x}| = - \frac{2}{\cos x} [∵ \frac{π}{2} < x < π and in the second quadrant, \cos x is negative] = RHS Hence proved .$

Page No 5.19:

Question 26:

If $T_{n} = \sin^{n} x + \cos^{n} x$ , prove that

(i) $\frac{T_{3} - T_{5}}{T_{1}} = \frac{T_{5} - T_{7}}{T_{3}}$

(ii) $2 T_{6} - 3 T_{4} + 1 = 0$

(iii) $6 T_{10} - 15 T_{8} + 10 T_{6} - 1 = 0$

Answer:

(i) LHS:

$\frac{T_{3} - T_{5}}{T_{1}} = \frac{(\sin^{3} x + \cos^{3} x) - (\sin^{5} x + \cos^{5} x)}{\sin x + \cos x} = \frac{\sin^{3} x - \sin^{5} x + \cos^{3} x - \cos^{5} x}{\sin x + \cos x} = \frac{\sin^{3} x (1 - \sin^{2} x) + \cos^{3} x (1 - \cos^{2} x)}{\sin x + \cos x} = \frac{\sin^{3} x . \cos^{2} x + c {os}^{3} x . \sin^{2} x}{\sin x + \cos x} = \frac{\sin^{2} x . \cos^{2} x (\sin x + c os x)}{\sin x + \cos x} = \sin^{2} x . \cos^{2} x$

RHS:
$\frac{T_{5} - T_{7}}{T_{3}} = \frac{(\sin^{5} x + \cos^{5} x) - (\sin^{7} x + \cos^{7} x)}{\sin^{3} x + \cos^{3} x} = \frac{\sin^{5} x - si n^{7} x + \cos^{5} x - \cos^{7} x}{\sin^{3} x + \cos^{3} x} = \frac{\sin^{5} x (1 - \sin^{2} x) + \cos^{5} x (1 - \cos^{2} x)}{\sin^{3} x + \cos^{3} x} = \frac{\sin^{5} x \cos^{2} x + \cos^{5} x \sin^{2} x}{\sin^{3} x + \cos^{3} x} = \sin^{2} x . \cos^{2} x$

LHS = RHS

Hence proved.

(ii) LHS:

$2 T_{6} - 3 T_{4} + 1 2 (\sin^{6} x + \cos^{6} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 2 (\sin^{2} x + \cos^{2} x) (\sin^{4} x + \cos^{4} x - \sin^{2} x \cos^{2} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 2.1 . (\sin^{4} x + \cos^{4} x - \sin^{2} x \cos^{2} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 2 \sin^{4} x + 2 \cos^{4} x - 2 \sin^{2} x \cos^{2} x - 3 \sin^{4} x - 3 \cos^{4} x + 1 - (\sin^{4} x + \cos^{4} x) - \sin^{2} x \cos^{2} x + 1 - (\sin^{2} x + \cos^{2} x)^{2} + 1 - 1 + 1 0$

Hence proved.

(iii) LHS:

$6 T_{10} - 15 T_{8} + 10 T_{6} - 1 6 (\sin^{10} x + \cos^{10} x) - 15 (\sin^{8} x + \cos^{8} x) + 10 (\sin^{6} x + \cos^{6} x) - 1$

Page No 5.25:

Question 1:

Find the values of the other five trigonometric functions in each of the following:

(i) $\cot x = \frac{12}{5},$ x in quadrant III

(ii) $\cos x = - \frac{1}{2},$ x in quadrant II

(iii) $\tan x = \frac{3}{4},$ x in quadrant III

(iv) $\sin x = \frac{3}{5},$ x in quadrant I

Answer:

$(i) We have : \cot x = \frac{12}{5} and x are in the third quadrant. In the third quadrant, \tan x and \cot x are positive A nd, \sin x, \cos x, \sec x and cosec x are negative .$
$∴ \tan x = \frac{1}{\cot x} = \frac{1}{\frac{12}{5}} = \frac{5}{12} cosec x = - \sqrt{1 + c o t^{2} x} = - \sqrt{1 + {(\frac{12}{5})}^{2}} = - \frac{13}{5} \sin x = \frac{1}{cosec x} = \frac{1}{- \frac{13}{5}} = - \frac{5}{13} \cos x = - \sqrt{1 - \sin^{2} x} = - \sqrt{1 - {(\frac{- 5}{13})}^{2}} = \frac{- 12}{13} And, \sec x = \frac{1}{\cos x} = \frac{1}{\frac{- 12}{13}} = \frac{- 13}{12}$
$(ii) We have : \cos x = - \frac{1}{2} and x are in the second quadrant. In the second quadrant, \sin x, and cosec x are positive . And, \tan x, \cot x, \cos x and \sec x are negative .$

$∴ \sin x = \sqrt{1 - \cos^{2} x} = \sqrt{1 - {(\frac{- 1}{2})}^{2}} = \frac{\sqrt{3}}{2} \tan x = \frac{\sin x}{\cos x} = \frac{\frac{\sqrt{3}}{2}}{- \frac{1}{2}} = - \sqrt{3}$
$c o t x = \frac{1}{\tan x} = \frac{1}{- \sqrt{3}} = \frac{- 1}{\sqrt{3}} \sec x = \frac{1}{\cos x} = \frac{1}{\frac{- 1}{2}} = - 2 cosec x = \frac{1}{\sin x} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$
$(iii) We have : \tan x = \frac{3}{4} and x are in the third quadrant. In the third quadrant, \tan x, and \cot x are positive . And, \sin x, \cos x, \sec x and cosec x are negative .$
$∴ c o t x = \frac{1}{\tan x} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \sec x = - \sqrt{1 + \tan^{2} x} = - \sqrt{1 + {(\frac{3}{4})}^{2}} = - \frac{5}{4} \cos x = \frac{1}{\sec x} = \frac{1}{- \frac{5}{4}} = - \frac{4}{5} \sin x = - \sqrt{1 - \cos^{2} x} = - \sqrt{1 - {(\frac{- 4}{5})}^{2}} = \frac{- 3}{5} cosec x = \frac{1}{\sin x} = \frac{1}{- \frac{3}{5}} = - \frac{5}{3}$
$(iv) We have : \sin x = \frac{3}{5} and x are in the first quadrant. In the first quadrant, all six T - ratios are positive .$
$∴ \cos x = \sqrt{1 - \sin^{2} x} = \sqrt{1 - {(\frac{3}{5})}^{2}} = \frac{4}{5} \tan x = \frac{\sin x}{\cos x} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \cot x = \frac{1}{\tan x} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \sec x = \frac{1}{\cos x} = \frac{1}{\frac{4}{5}} = \frac{5}{4} cosec x = \frac{1}{\sin x} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$

Page No 5.25:

Question 2:

If sin $x = \frac{12}{13}$ and x lies in the second quadrant, find the value of sec x + tan x.

Answer:

$We have : \sin x = \frac{12}{13} and x lie in the second quadrant . In the second quadrant, \sin x and cosec x are positive and all the other four T - ratios are negative .$
$∴ \cos x = - \sqrt{1 - \sin^{2} x} = - \sqrt{1 - {(\frac{12}{13})}^{2}} = \frac{- 5}{13} \tan x = \frac{\sin x}{\cos x} = \frac{\frac{12}{13}}{\frac{- 5}{13}} = \frac{- 12}{5} And, sec x = \frac{1}{\cos x} = \frac{1}{\frac{- 5}{13}} = \frac{- 13}{5} ∴ sec x + \tan x = \frac{- 13}{5} + \frac{- 12}{5} = - 5$

Page No 5.25:

Question 3:

If sin $x = \frac{3}{5}, \tan y = \frac{1}{2} and \frac{π}{2} < x < π < y < \frac{3 π}{2},$ find the value of 8 tan $x - \sqrt{5} \sec y$ .

Answer:

We have:
$\sin x = \frac{3}{5}, \tan y = \frac{1}{2} and \frac{π}{2} < x < π < y < \frac{3 π}{2}, Thus, x is in the second quadrant and y is in the third quadrant . In the second quadrant, \cos x and \tan x are negative . In the third quadrant, \sec y is negative .$
$∴ \cos x = - \sqrt{1 - \sin^{2} x} = - \sqrt{1 - {(\frac{3}{5})}^{2}} = \frac{- 4}{5} \tan x = \frac{\frac{3}{5}}{\frac{- 4}{5}} = \frac{- 3}{4} And, \sec y = - \sqrt{1 + \tan^{2} y} = - \sqrt{1 + {(\frac{1}{2})}^{2}} = \frac{- \sqrt{5}}{2} ∴ 8 \tan x - \sqrt{5} \sec y = 8 \times \frac{- 3}{4} - \sqrt{5} \times \frac{- \sqrt{5}}{2} = - 6 + \frac{5}{2} = - \frac{7}{2}$

Page No 5.25:

Question 4:

If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.

Answer:

$We have : \sin x + \cos x = 0 \Rightarrow \sin x = - \cos x \Rightarrow \frac{\sin x}{\cos x} = - 1 \Rightarrow \tan x = - 1$

$Now, x is in the fourth quadrant . In the fourth quadrant, \cos x and \sec x are positive and all the other four T - ratios are negative .$

$∴ s e c x = \sqrt{1 + \tan^{2} x} = \sqrt{1 + {(- 1)}^{2}} = \sqrt{2} \cos x = \frac{1}{s e c x} = \frac{1}{\sqrt{2}} And, \sin x = - \sqrt{1 - \cos^{2} x} = - \sqrt{1 - {(\frac{1}{\sqrt{2}})}^{2}} = \frac{- 1}{\sqrt{2}} ∴ \sin x = \frac{- 1}{\sqrt{2}} and \cos x = \frac{1}{\sqrt{2}}$

Page No 5.25:

Question 5:

If $\cos x = - \frac{3}{5} and π < x < \frac{3 π}{2}$ find the values of other five trigonometric functions and hence evaluate $\frac{cosec x + \cot x}{\sec x - \tan x}$ .

Answer:

$We have : \cos x = - \frac{3}{5} and π < x < \frac{3 π}{2} Thus, x is in the third quadrant . In the third quadrant, \tan x and \cot x are positive And, all the other four T - ratios are negative . ∴ \sin x = - \sqrt{1 - \cos^{2} x} = - \sqrt{1 - {(\frac{- 3}{5})}^{2}} = \frac{- 4}{5} \tan x = \frac{\sin x}{\cos x} = \frac{\frac{- 4}{5}}{- \frac{3}{5}} = \frac{4}{3} \cot x = \frac{1}{\tan x} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \sec x = \frac{1}{\cos x} = \frac{1}{\frac{- 3}{5}} = \frac{- 5}{3} cosec x = \frac{1}{\sin x} = \frac{1}{\frac{- 4}{5}} = \frac{- 5}{4}$

$Now, \frac{\cos e c θ + cotθ}{s e c θ - tanθ} = \frac{\frac{- 5}{4} + \frac{3}{4}}{\frac{- 5}{3} - \frac{4}{3}} = \frac{\frac{- 2}{4}}{\frac{- 9}{3}} = \frac{\frac{- 1}{2}}{- 3} = \frac{1}{6}$

Page No 5.39:

Question 1:

Find the values of the following trigonometric ratios:
(i) $\sin \frac{5 π}{3}$

(ii) sin 17π

(iii) $\tan \frac{11 π}{6}$

(iv) $\cos (- \frac{25 π}{4})$

(v) $\tan \frac{7 π}{4}$

(vi) $\sin \frac{17 π}{6}$

(vii) $\cos \frac{19 π}{6}$

(viii) $\sin (- \frac{11 π}{6})$

(ix) $cosec (- \frac{20 π}{3})$

(x) $\tan (- \frac{13 π}{4})$

(xi) $\cos \frac{19 π}{4}$

(xii) $\sin \frac{41 π}{4}$

(xiii) $\cos \frac{39 π}{4}$

(xiv) $\sin \frac{151 π}{6}$

Answer:

$(i) We have : \frac{5 π}{3} = {(\frac{5}{3} \times 180)}^{°} = 300 ° = (90 ° \times 3 + 30 °) 300 ° lies in the fourth quadrant in which the sine function is negative . Also, 3 is an odd integer . ∴ \sin (\frac{5 π}{3}) = \sin (300 °) = \sin (90 ° \times 3 + 30 °) = - \cos 30 ° = - \frac{\sqrt{3}}{2}$

$(ii) We have : \sin 17 π = \sin 3060 ° 3060 ° = 90 ° \times 34 + 0 ° Clearly 3060 ° is in the negative direction of the x - axis, i . e . on the boundary line of the II and III quadrant s . Also, 34 is an even integer . ∴ \sin (3060 °) = \sin (90 ° \times 34 + 0 °) = - \sin 0 ° = 0$

$(iii) We have : \frac{11 π}{6} = {(\frac{11}{6} \times 180)}^{°} = 330 ° = (90 ° \times 3 + 60 °) 330 ° lies in the fourth quadrant in which the tangent function is negative . Also, 3 is an odd integer . ∴ \tan (\frac{11 π}{6}) = \tan (330 °) = \tan (90 ° \times 3 + 60 °) = - \cot 60 ° = - \frac{1}{\sqrt{3}}$

$(iv) We have : \cos (- \frac{25 π}{4}) = \cos (1125 °) \cos (- 1125 °) = \cos (1125 °) = \cos (90 ° \times 12 + 45 °) 1125 ° lies in the first quadrant in which the cosine function is positive . Also, 12 is an even integer . ∴ \cos (- 1125 °) = \cos (1125 °) = \cos (90 ° \times 12 + 45 °) = \cos 45 ° = \frac{1}{\sqrt{2}}$

$(v) We have : \tan \frac{7 π}{4} = \tan 315 ° 315 ° = (90 ° \times 3 + 45 °) 315 ° lies in the fourth quadrant in which the tangent function is negative . Also, 3 is an odd integer . ∴ \tan (315 °) = \tan (90 ° \times 3 + 45 °) = - \cot 45 ° = - 1$

$(vi) We have : \sin \frac{17 π}{6} = \sin 510 ° 510 ° = (90 ° \times 5 + 60) ° 510 ° lies in the second quadrant in which the sine function is positive . Also, 5 is an odd integer . ∴ \sin (510 °) = \sin (90 ° \times 5 + 60 °) = \cos (60 °) = \frac{1}{2}$

$(vii) We have : \cos \frac{19 π}{6} = \cos 570 ° 570 ° = (90 ° \times 6 + 30 °) 570 ° lies in the third quadrant in which the cosine function is negative . Also, 6 is an odd integer . ∴ \cos (570 °) = \cos (90 ° \times 6 + 30 °) = - \cos (30 °) = - \frac{\sqrt{3}}{2}$

$(viii) We have : \sin (- \frac{11 π}{6}) = \sin (- 330 °) \sin (- 330 °) = - \sin (330 °) = - \sin (90 ° \times 3 + 60 °) 330 ° lies in the fourth quadrant in which the sine function is negative . Also, 3 is an odd integer . ∴ \sin (- 330 °) = - \sin (330 °) = - \sin (90 ° \times 3 + 60 °) = - (- \cos 60 °) = - (- \frac{1}{2}) = \frac{1}{2}$

$(ix) We have : cosec (- \frac{20 π}{3}) = cosec (- 1200 °) cosec (- 1200 °) = - cosec (1200 °) = - cosec (90 ° \times 13 + 30 °) 1200 ° lies in the second quadrant in which the cosec function is positive . Also, 13 is an odd integer . ∴ cosec (- 1200 °) = - cosec (1200 °) = - cosec (90 ° \times 13 + 30 °) = - \sec 30 ° = - \frac{2}{\sqrt{3}}$

$(x) We have : \tan (- \frac{13 π}{4}) = \tan (- 585 °) \tan (- 585 °) = - \tan (585 °) = - \tan (90 ° \times 6 + 45 °) 585 ° lies in the third quadrant in which tangent function is positive . Also, 6 is an even integer . ∴ \tan (- 585 °) = - \tan (585 °) = - \tan (90 ° \times 6 + 45 °) = - \tan 45 ° = - 1$

$(xi) We have : \cos \frac{19 π}{4} = \cos 855 ° 855 ° = 90 ° \times 9 + 45 ° 855 ° lies in the second quadrant in which the cosine function is negative . Also, 9 is an odd integer . ∴ \cos (855 °) = \cos (90 ° \times 9 + 45 °) = - \sin (45 °) = - \frac{1}{\sqrt{2}}$

$(xii) We have : \sin \frac{41 π}{4} = \sin 1845 ° 1845 ° = 90 ° \times 20 + 45 ° 1845 ° lies in the first quadrant in which the sine function is positive . Also, 20 is an even integer . ∴ \sin (1845 °) = \sin (90 ° \times 20 + 45 °) = \sin (45 °) = \frac{1}{\sqrt{2}}$

$(xiii) We have : \cos \frac{39 π}{4} = \cos 1755 ° 1755 ° = 90 ° \times 19 + 45 ° 1755 ° lies in the fourth quadrant in which the cosine function is positive . Also, 19 is an odd integer . ∴ \cos (1755 °) = \cos (90 ° \times 19 + 45 °) = \sin (45 °) = \frac{1}{\sqrt{2}}$

$(xiv) We have : \sin \frac{151 π}{6} = \sin 4530 ° 4530 ° = 90 ° \times 50 + 30 ° 4530 ° lies in the third quadrant in which the sine function is negative . Also, 50 is an even integer . ∴ \sin (4530 °) = \sin (90 ° \times 50 + 30 °) = - \sin (30 °) = - \frac{1}{2}$

Page No 5.39:

Question 2:

Prove that:
(i) tan 225° cot 405° + tan 765° cot 675° = 0
(ii) $\sin \frac{8 π}{3} \cos \frac{23 π}{6} + \cos \frac{13 π}{3} \sin \frac{35 π}{6} = \frac{1}{2}$
(iii) cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = $\frac{1}{2}$
(iv) tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
(v) cos 570° sin 510° + sin (−330°) cos (−390°) = 0
(vi) $\tan \frac{11 π}{3} - 2 \sin \frac{4 π}{6} - \frac{3}{4} {cosec}^{2} \frac{π}{4} + 4 \cos^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}$
(vii) $3 \sin \frac{π}{6} \sec \frac{π}{3} - 4 \sin \frac{5 π}{6} \cot \frac{π}{4} = 1$

Answer:

$(i) LHS = \tan 225 ° \cot 405 ° + \tan 765 ° \cot 675 ° = \tan (90 ° \times 2 + 45 °) \cot (90 ° \times 4 + 45 °) + \tan (90 ° \times 8 + 45 °) \cot (90 ° \times 7 + 45 °) = \tan (45 °) \cot (45 °) + \tan (45 °) [- \tan (45 °)] = 1 \times 1 + 1 \times (- 1) = 1 - 1 = 0 = RHS Hence proved .$

$(ii) LHS = \sin \frac{8 π}{3} \cos \frac{23 π}{6} + \cos \frac{13 π}{3} \sin \frac{35 π}{6} = \sin (\frac{8}{3} \times 180 °) \cos (\frac{23}{6} \times 180 °) + \cos (\frac{13}{3} \times 180 °) \sin (\frac{35}{6} \times 180 °) = \sin (480 °) \cos (690 °) + \cos (780 °) \sin (1050 °) = \sin (90 ° \times 5 + 30 °) \cos (90 ° \times 7 + 60 °) + \cos (90 ° \times 8 + 60 °) \sin (90 ° \times 11 + 60 °) = \cos (30 °) \sin (60 °) + \cos (60 °) [- \cos (60 °)] = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times (- \frac{1}{2}) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} = RHS Hence proved .$

$(iii) LHS = \cos 24 ° + \cos 55 ° + \cos 125 ° + \cos 204 ° + \cos 300 ° = \cos 24 ° + \cos (90 ° - 35 °) + \cos (90 ° \times 1 + 35 °) + \cos (90 ° \times 2 + 24 °) + \cos (90 ° \times 3 + 30 °) = \cos 24 ° + \sin 35 ° - \sin 35 ° - \cos 24 ° + \sin 30 ° = 0 + 0 + \frac{1}{2} = \frac{1}{2} = RHS Hence proved .$

$(iv) LHS = \tan (- 225 °) \cot (- 405 °) - \tan (- 765 °) \cot (675 °) = [- \tan (225 °)] [- \cot (405 °)] - [- \tan (765 °)] \cot (675 °) [∵ \tan (- x) = \tan (x) and \cot (- x) = - \cot (x)] = \tan (225 °) \cot (405 °) + \tan (765 °) \cot (675 °) = \tan (90 ° \times 2 + 45 °) \cot (90 ° \times 4 + 45 °) + \tan (90 ° \times 8 + 45 °) \cot (90 ° \times 7 + 45 °) = \tan (45 °) \cot (45 °) + \tan (45 °) [- \tan (45 °)] = 1 \times 1 + 1 \times (- 1) = 1 - 1 = 0 = RHS Hence, proved .$

$(v) LHS = \cos (570 °) \sin (510 °) + \sin (- 330 °) \cos (- 390 °) = \cos (570 °) \sin (510 °) + [- \sin (330 °)] \cos (390 °) [∵ \sin (- x) = - \sin x and \cos (- x) = \cos x] = \cos (570 °) \sin (510 °) - \sin (330 °) \cos (390 °) = \cos (90 ° \times 6 + 30 °) \sin (90 ° \times 5 + 60 °) - \sin (90 ° \times 3 + 60 °) \cos (90 ° \times 4 + 30 °) = - \cos (30 °) \cos (60 °) - [- \cos (60 °)] \cos (30 °) = - \cos (30 °) \cos (60 °) + \cos (30 °) \sin (60 °) = 0 = RHS Hence proved .$

$(vi) LHS = \tan \frac{11 π}{3} - 2 \sin \frac{4 π}{6} - \frac{3}{4} \cos e c^{2} \frac{π}{4} + 4 \cos^{2} \frac{17 π}{6} = \tan (\frac{11 π}{3}) - 2 \sin (\frac{4 π}{6}) - \frac{3}{4} {[\cos e c (\frac{π}{4})]}^{2} + 4 {[\cos (\frac{17 π}{6})]}^{2} = \tan (\frac{11}{3} \times 180 °) - 2 \sin (\frac{4}{6} \times 180 °) - \frac{3}{4} {[\cos e c (\frac{180 °}{4})]}^{2} + 4 {[\cos (\frac{17 \times 180 °}{6})]}^{2} = \tan (660 °) - 2 \sin (120 °) - \frac{3}{4} {[\cos e c (45 °)]}^{2} + 4 {[\cos (510 °)]}^{2} = \tan (660 °) - 2 \sin (120 °) - \frac{3}{4} {[cosec (45 °)]}^{2} + 4 {[\cos (510 °)]}^{2} = \tan (90 ° \times 7 + 30 °) - 2 \sin (90 ° \times 1 + 30 °) - \frac{3}{4} {[cosec (45 °)]}^{2} + 4 {[\cos (90 ° \times 5 + 60 °)]}^{2} = [- \cot (30 °)] - [2 \cos (30 °)] - \frac{3}{4} {[cosec (45 °)]}^{2} + 4 {[- \sin (60 °)]}^{2} = - \cot (30 °) - 2 \cos (30 °) - \frac{3}{4} {[cosec (45 °)]}^{2} + 4 {[\sin (60 °)]}^{2} = - \sqrt{3} - \frac{2 \sqrt{3}}{2} - \frac{3}{4} {[\sqrt{2}]}^{2} + 4 {[\frac{\sqrt{3}}{2}]}^{2} = - \sqrt{3} - \sqrt{3} - \frac{3}{2} + 3 = \frac{3 - 4 \sqrt{3}}{2} = RHS Hence proved .$

$(v ii) LHS = 3 \sin \frac{π}{6} s e c \frac{π}{3} - 4 \sin \frac{5 π}{6} c o t \frac{π}{4} = 3 \sin (\frac{180 °}{6}) \sec (\frac{180 °}{3}) - 4 \sin (\frac{5 \times 180 °}{6}) \cot (\frac{180 °}{4}) = 3 \sin (30 °) \sec (60 °) - 4 \sin (150 °) \cot (45 °) = 3 \sin (30 °) \sec (60 °) - 4 \sin (90 ° \times 1 + 60 °) \cot (45 °) = 3 \sin (30 °) \sec (60 °) - 4 \cos (60 °) \cot (45 °) = 3 \times \frac{1}{2} \times 2 - 4 \times \frac{1}{2} \times 1 = 3 - 2 = 1 = RHS Hence proved .$

Page No 5.39:

Question 3:

Prove that

(i) $\frac{\cos (2 π + x) cosec (2 π + x) \tan (π / 2 + x)}{\sec (π / 2 + x) \cos x \cot (π + x)} = 1$

(ii) $\frac{cosec (90 ° + x) + \cot (450 ° + x)}{cosec (90 ° - x) + \tan (180 ° - x)} + \frac{\tan (180 ° + x) + \sec (180 ° - x)}{\tan (360 ° + x) - \sec (- x)} = 2$

(iii) $\frac{\sin (180 ° + x) \cos (90 ° + x) \tan (270 ° - x) \cot (360 ° - x)}{\sin (360 ° - x) \cos (360 ° + x) cosec (- x) \sin (270 ° + x)} = 1$

(iv) $\{1 + \cot x - \sec (\frac{π}{2} + x)\} \{1 + \cot x + \sec (\frac{π}{2} + x)\} = 2 \cot x$

(v) $\frac{\tan (90 ° - x) \sec (180 ° - x) \sin (- x)}{\sin (180 ° + x) \cot (360 ° - x) cosec (90 ° - x)} = 1$

Answer:

$(i) LHS = \frac{\cos (2 π + x) \cos e c (2 π + x) \tan (\frac{π}{2} + x)}{\sec (\frac{π}{2} + x) \cos x \cot (π + x)} = \frac{\cos x cosec x [- \cot x]}{[- \cos e c x] \cos x \cot x} = \frac{- \cos x cosec x \cot x}{- cosec x c os x \cot x} = 1 = RHS Hence proved .$

$(ii) LHS = \frac{cosec (90 ° + x) + \cot (450 ° + x)}{cosec (90 ° - x) + \tan (180 ° - x)} + \frac{\tan (180 ° + x) + \sec (180 ° - x)}{\tan (360 ° + x) - \sec (- x)} = \frac{cosec (90 ° + x) + \cot (450 ° + x)}{cosec (90 ° - x) + \tan (180 ° - x)} + \frac{\tan (180 ° + x) + \sec (180 ° - x)}{\tan (360 ° + x) - \sec (- x)} = \frac{cosec (90 ° + x) + \cot (90 ° \times 5 + x)}{cosec (90 ° - x) + \tan (90 ° \times 2 - x)} + \frac{\tan (90 ° \times 2 + x) + \sec (90 ° \times 2 - x)}{\tan (90 ° \times 4 + x) - \sec (- x)} = \frac{\sec x + \cot (90 ° \times 5 + x)}{cosec (90 ° - x) + \tan (90 ° \times 2 - x)} + \frac{\tan (90 ° \times 2 + x) + \sec (90 ° \times 2 - x)}{\tan (90 ° \times 4 + x) - \sec (- x)} = \frac{\sec x - \tan x}{\sec x - \tan x} + \frac{\tan x - \sec x}{\tan x - \sec x} = 1 + 1 = 2 = RHS Hence proved .$
$(iii) LHS = \frac{\sin (180 ° + x) \cos (90 ° + x) \tan (270 ° - x) \cot (360 ° - x)}{\sin (360 ° - x) \cos (360 ° + x) cosec (- x) \sin (270 ° + x)} = \frac{\sin (90 \times 2 ° + x) \cos (90 ° \times 1 + x) \tan (90 ° \times 3 - x) \cot (90 ° \times 4 - x)}{\sin (90 ° \times 4 - x) \cos (90 ° \times 4 + x) cosec (- x) \sin (90 ° \times 3 + x)} = \frac{- \sin x [- \sin x] \cot x [- \cot x]}{[- \sin x] \cos x [- \cos e c x] [- \cos x]} = \frac{\sin^{2} x \cot^{2} x}{\sin x \cos e c x \cos x \cos x} = \frac{\sin^{2} x \times \frac{\cos^{2} x}{\sin^{2} x}}{\sin x \times \frac{1}{\sin x} \times \cos^{2} x} = \frac{\cos^{2} x}{\cos^{2} x} = 1 = RHS Hence proved .$

$(iv) LHS = \{1 + \cot x - \sec (\frac{π}{2} + x)\} \{1 + \cot x + \sec (\frac{π}{2} + x)\} = [1 + \cot x - \{- cosec x\}] [1 + \cot x + \{- cosec x\}] = [1 + \cot x + cosec x] [1 + \cot x - cosec x] = [1 + \cot x + cosec x] [1 + \cot x - cosec x] = [\{1 + \cot (x)\} + \{cosec x\}] [\{1 + \cot x\} - \{cosec x\}] = {\{1 + \cot x\}}^{2} - {\{cosec x\}}^{2}$

$= 1 + \cot^{2} x + 2 \cot x - {cosec}^{2} x = 2 \cot x [∵ 1 + \cot^{2} x = {cosec}^{2} x] = RHS Hence proved .$

$(v) LHS = \frac{\tan (90 ° - x) \sec (180 ° - x) \sin (- x)}{\sin (180 ° + x) \cot (360 ° - x) cosec (90 ° - x)} = \frac{\tan (90 ° \times 1 - x) \sec (90 ° \times 2 - x) \sin (- x)}{\sin (90 ° \times 2 + x) \cot (90 ° \times 4 - x) cosec (90 ° \times 1 - x)} = \frac{\cot x [- \sec x] [- \sin x]}{[- \sin x] [- \cot x] \sec x} = \frac{\cot x \sec x \sin x}{\sin x \cot x \sec x} = \frac{\frac{\cos x}{\sin x} \times \frac{1}{\cos x} \times \sin x}{\sin x \times \frac{\cos x}{\sin x} \times \frac{1}{\cos x}} = \frac{1}{1} = 1 = RHS Hence proved .$

Page No 5.40:

Question 4:

Prove that: $\sin^{2} \frac{π}{18} + \sin^{2} \frac{π}{9} + \sin^{2} \frac{7 π}{18} + \sin^{2} \frac{4 π}{9} = 2$

Answer:

$LHS = \sin^{2} \frac{π}{18} + \sin^{2} \frac{π}{9} + \sin^{2} \frac{7 π}{18} + \sin^{2} \frac{4 π}{9} = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \sin^{2} \frac{7 π}{18} + \sin^{2} \frac{8 π}{18} = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \sin^{2} (\frac{7 π}{18}) + \sin^{2} (\frac{8 π}{18}) = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \sin^{2} (\frac{π}{2} - \frac{2 π}{18}) + \sin^{2} (\frac{π}{2} - \frac{π}{18}) = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \cos^{2} \frac{2 π}{18} + \cos^{2} \frac{π}{18} = \sin^{2} \frac{π}{18} + \cos^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \cos^{2} \frac{2 π}{18} = 1 + 1 = 2 = RHS Hence proved .$

Page No 5.40:

Question 5:

Prove that:
$\sec (\frac{3 π}{2} - x) \sec (x - \frac{5 π}{2}) + \tan (\frac{5 π}{2} + x) \tan (x - \frac{3 π}{2}) = - 1 .$

Answer:

$LHS = \sec (\frac{3 π}{2} - x) \sec (x - \frac{5 π}{2}) + \tan (\frac{5 π}{2} + x) \tan (x - \frac{3 π}{2}) = \sec (\frac{3 π}{2} - x) \sec [- (\frac{5 π}{2} - x)] + \tan (\frac{5 π}{2} + x) \tan [- (\frac{3 π}{2} - x)] = \sec (\frac{3 π}{2} - x) \sec (\frac{5 π}{2} - x) + \tan (\frac{5 π}{2} + x) [- \tan (\frac{3 π}{2} - x)] = \sec (\frac{3 π}{2} - x) \sec (\frac{5 π}{2} - x) - \tan (\frac{5 π}{2} + x) \tan (\frac{3 π}{2} - x) = \sec (\frac{π}{2} \times 3 - x) \sec (\frac{π}{2} \times 5 - x) - \tan (\frac{π}{2} \times 5 + x) \tan (\frac{π}{2} \times 3 - x) = [- cosec x] [cosec x] - [- \cot x] \cot x = - {cosec}^{2} x + \cot^{2} x = - [{cosec}^{2} x - \cot^{2} x] = - 1 = RHS Hence proved .$

Page No 5.40:

Question 6:

In a âˆ†ABC, prove that:
(i) cos (A + B) + cos C = 0
(ii) $\cos (\frac{A + B}{2}) = \sin \frac{C}{2}$
(iii) $\tan \frac{A + B}{2} = \cot \frac{C}{2}$

Answer:

$(i) In ∆ A B C : A + B + C = π ∴ A + B = π - C Now, LHS = \cos (A + B) + \cos C = \cos (π - C) + \cos C = - \cos (C) + \cos C [∵ \cos (π - C) = - \cos (C)] = 0 = RHS H ence proved .$

$(ii) In ∆ A B C : A + B + C = π \Rightarrow A + B = π - C \Rightarrow \frac{A + B}{2} = \frac{π - C}{2} \Rightarrow \frac{A + B}{2} = \frac{π}{2} - \frac{C}{2} Now, LHS = \cos (\frac{A + B}{2}) = \cos (\frac{π}{2} - \frac{C}{2}) = \sin (\frac{C}{2}) [∵ \cos (\frac{π}{2} - θ) = \sin θ] = RHS Hence proved .$

$(iii) In ∆ A B C : A + B + C = π \Rightarrow A + B = π - C \Rightarrow \frac{A + B}{2} = \frac{π - C}{2} \Rightarrow \frac{A + B}{2} = \frac{π}{2} - \frac{C}{2} Now, LHS = \tan (\frac{A + B}{2}) = \tan (\frac{π}{2} - \frac{C}{2}) = \cot (\frac{C}{2}) [∵ \tan (\frac{π}{2} - θ) = \cot θ] = RHS Hence proved .$

Page No 5.40:

Question 7:

In a âˆ†A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that
cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0

Answer:

$A, B, C and D are the angles of a cyclic quadrilateral . ∴ A + C = 180 ° and B + D = 180 ° \Rightarrow A = 180 - C and B = 180 - D Now, LHS = \cos (180 ° - A) + \cos (180 ° + B) + \cos (180 ° + C) - \sin (90 ° + D) = - \cos A + [- \cos B] + [- \cos C] - \cos D = - \cos A - \cos B - \cos C - \cos D = - \cos (180 ° - C) - \cos (180 ° - D) - \cos C - \cos D = - [- \cos C] - [- \cos D] - \cos C - \cos D = \cos C + \cos D - \cos C - \cos D = 0 = RHS Hence proved .$

Page No 5.40:

Question 8:

Find x from the following equations:
$(i) cosec (\frac{π}{2} + θ) + x \cos θ \cot (\frac{π}{2} + θ) = \sin (\frac{π}{2} + θ) (ii) x \cot (\frac{π}{2} + θ) + \tan (\frac{π}{2} + θ) \sin θ + cosec (\frac{π}{2} + θ) = 0$

Answer:

$90 ° = \frac{π}{2}$
$(i) We have : cosec (90 ° + θ) + x \cos θ \cot (90 ° + θ) = \sin (90 ° + θ) \Rightarrow \sec θ + x \cos θ [- \tan θ] = \cos θ \Rightarrow \sec θ - x cosθ tanθ = \cos θ \Rightarrow \sec θ - x cosθ \times \frac{\sin θ}{\cos θ} = \cos θ \Rightarrow \sec θ - x \sin θ = \cos θ \Rightarrow \sec θ - \cos θ = x \sin θ \Rightarrow \frac{1}{\cos θ} - cosθ = x \sin θ \Rightarrow \frac{1 - \cos^{2} θ}{\cos θ} = x \sin θ \Rightarrow \frac{\sin^{2} θ}{cosθ} = x \sin θ \Rightarrow \frac{\sin^{2} θ}{\cos θ \sin θ} = x \Rightarrow \frac{\sin θ}{\cos θ} = x \Rightarrow \tan θ = x ∴ x = \tan θ$

$(ii) We have : x \cot (90 ° + θ) + \tan (90 ° + θ) \sin θ + cosec (90 ° + θ) = 0 \Rightarrow x [- \tan θ] + [- \cot θ] \sin θ + \sec θ = 0 \Rightarrow - x \tan θ - \cot θ \sin θ + \sec θ = 0 \Rightarrow - x \times \frac{\sin θ}{\cos θ} - \frac{\cos θ}{\sin θ} \times \sin θ + \frac{1}{\cos θ} = 0 \Rightarrow - x \times \frac{\sin θ}{\cos θ} - \cos θ + \frac{1}{\cos θ} = 0 \Rightarrow \frac{- x \sin θ - \cos^{2} θ + 1}{\cos θ} = 0$
^{$\Rightarrow - x \sin θ - \cos^{2} θ + 1 = 0 \Rightarrow - x \sin θ + \sin^{2} θ = 0 \Rightarrow x \sin θ = \sin^{2} θ \Rightarrow x = \frac{\sin^{2} θ}{\sin θ} \Rightarrow x = \sin θ$}

Page No 5.40:

Question 9:

Prove that:
(i) $\tan 4 π - \cos \frac{3 π}{2} - \sin \frac{5 π}{6} \cos \frac{2 π}{3} = \frac{1}{4}$

(ii) $\sin \frac{13 π}{3} \sin \frac{8 π}{3} + \cos \frac{2 π}{3} \sin \frac{5 π}{6} = \frac{1}{2}$

(iii) $\sin \frac{13 π}{3} \sin \frac{2 π}{3} + \cos \frac{4 π}{3} \sin \frac{13 π}{6} = \frac{1}{2}$

(iv) $\sin \frac{10 π}{3} \cos \frac{13 π}{6} + \cos \frac{8 π}{3} \sin \frac{5 π}{6} = - 1$

(v) $\tan \frac{5 π}{4} \cot \frac{9 π}{4} + \tan \frac{17 π}{4} \cot \frac{15 π}{4} = 0$

Answer:

$(i) 4 π = 720 °, \frac{3 π}{2} = 270 °, \frac{5 π}{6} = 150 °, \frac{2 π}{3} = 120 ° LHS = \tan (720 °) - \cos (270 °) - \sin (150 °) \cos (120 °) = \tan (90 ° \times 8 + 0 °) - \cos (90 ° \times 3 + 0 °) - \sin (90 ° \times 1 + 60 °) \cos (90 ° \times 1 + 30 °) = \tan (0 °) - \sin (0 °) - \cos (60 °) [- \sin (30 °)] = \tan (0 °) - \sin (0 °) + \cos (60 °) \sin (30 °) = 0 - 0 + \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = RHS Hence proved .$

$(ii) \frac{13 π}{3} = 780 °, \frac{8 π}{3} = 480 °, \frac{2 π}{3} = 120 °, \frac{5 π}{6} = 150 ° LHS = \sin (780 °) \sin (480 °) + \cos (120 °) \sin (150 °) = \sin (90 ° \times 8 + 60 °) \sin (90 ° \times 5 + 30 °) + \cos (90 ° \times 1 + 30 °) \sin (90 ° \times 1 + 60 °) = \sin (60 °) \cos (30 °) + [- \sin (30 °)] \cos (60 °) = \sin (60 °) \cos (30 °) - \sin (30 °) \cos (60 °) = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = RHS Hence proved .$

$(iii) \frac{13 π}{3} = 780 °, \frac{2 π}{3} = 120 °, \frac{4 π}{3} = 240 °, \frac{13 π}{6} = 390 ° LHS = \sin (780 °) \sin (120 °) + \cos (240 °) \sin (390 °) = \sin (90 ° \times 8 + 60 °) \sin (90 ° \times 1 + 30 °) + \cos (90 ° \times 2 + 60 °) \sin (90 ° \times 4 + 30 °) = \sin 60 ° \cos 30 ° + [- \cos 60 °] \sin 30 ° = \sin 60 ° \cos 30 ° - \cos 60 ° \sin 30 ° = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = RHS Hence proved .$

$(iv) \frac{10 π}{3} = 600 °, \frac{13 π}{6} = 390 °, \frac{8 π}{3} = 480 °, \frac{5 π}{6} = 150 ° LHS = \sin 600 ° \cos 390 ° + \cos 480 ° \sin 150 ° = \sin (90 ° \times 6 + 60 °) \cos (90 ° \times 4 + 30 °) + \cos (90 ° \times 5 + 30 °) \sin (90 ° \times 1 + 60 °) = [- \sin 60 °] \cos 30 ° + [- \sin 30 °] \cos 60 ° = - \sin 60 ° \cos (30 °) - \sin 30 ° \cos 60 ° = - \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = - \frac{3}{4} - \frac{1}{4} = - 1 = RHS Hence proved .$

$(v) \frac{5 π}{4} = 225 °, \frac{9 π}{4} = 405 °, \frac{17 π}{4} = 765 °, \frac{15 π}{4} = 675 ° LHS = \tan 225 ° \cot 405 ° + \tan 765 ° \cot 675 ° = \tan (90 ° \times 2 + 45 °) \cot (90 ° \times 4 + 45 °) + \tan (90 ° \times 8 + 45 °) \cot (90 ° \times 7 + 45 °) = \tan 45 ° \cot 45 ° + \tan 45 ° [- \tan 45 °] = \tan 45 ° \cot 45 ° - \tan 45 ° \tan 45 ° = 1 \times 1 - 1 \times 1 = 1 - 1 = 0 = RHS Hence proved .$

Page No 5.40:

Question 1:

If tan x = $x - \frac{1}{4 x}$ , then sec x − tan x is equal to
(a) $- 2 x, \frac{1}{2 x}$
(b) $- \frac{1}{2 x}, 2 x$
(c) 2x
(d) $2 x, \frac{1}{2 x}$

Answer:

(a) $- 2 x, \frac{1}{2 x}$

$We have, \tan x = x - \frac{1}{4 x} \Rightarrow s e c^{2} x = 1 + \tan^{2} x \Rightarrow s e c^{2} x = 1 + {(x - \frac{1}{4 x})}^{2} \Rightarrow s e c^{2} x = x^{2} + \frac{1}{16 x^{2}} + \frac{1}{2} \Rightarrow s e c^{2} x = {(x + \frac{1}{4 x})}^{2} ∴ s e c x = \pm (x + \frac{1}{4 x}) \Rightarrow s e c x - \tan x = (x + \frac{1}{4 x}) - (x - \frac{1}{4 x}) or - (x + \frac{1}{4 x}) - (x - \frac{1}{4 x}) = \frac{1}{2 x} or - 2 x$

Page No 5.40:

Question 2:

If sec $x = x + \frac{1}{4 x}$ , then sec x + tan x =
(a) $x, \frac{1}{x}$
(b) $2 x, \frac{1}{2 x}$
(c) $- 2 x, \frac{1}{2 x}$
(d) $- \frac{1}{x}, x$

Answer:

(b) $2 x, \frac{1}{2 x}$

$We have, s e c x = x + \frac{1}{4 x} \Rightarrow s e c^{2} x = = x^{2} + \frac{1}{16 x^{2}} + \frac{1}{2} \Rightarrow 1 + \tan^{2} x = 1 + x^{2} + \frac{1}{16 x^{2}} - \frac{1}{2} \Rightarrow \tan^{2} x = x^{2} + \frac{1}{16 x^{2}} - \frac{1}{2} \Rightarrow \tan^{2} x = {(x - \frac{1}{4 x})}^{2} ∴ \tan x = \pm (x - \frac{1}{4 x}) \Rightarrow s e c x - \tan x = (x + \frac{1}{4 x}) - (x - \frac{1}{4 x}) or (x + \frac{1}{4 x}) - [- (x - \frac{1}{4 x})] = \frac{1}{2 x} or 2 x$

Page No 5.40:

Question 3:

If $\frac{π}{2} < x < \frac{3 π}{2}, then \sqrt{\frac{1 - \sin x}{1 + \sin x}}$ is equal to
(a) sec x − tan x
(b) sec x + tan x
(c) tan x − sec x
(d) none of these

Answer:

(c) tan x − sec x

$\sqrt{\frac{1 - \sin x}{1 + \sin x}} = \sqrt{\frac{(1 - \sin x) (1 - \sin x)}{(1 + \sin x) (1 - \sin x)}} = \sqrt{\frac{{(1 - \sin x)}^{2}}{1 - \sin^{2} x}} = \sqrt{\frac{{(1 - \sin x)}^{2}}{\cos^{2} x}} = \frac{(1 - \sin x)}{- co s x} [as, \frac{π}{2} < x < \frac{3 π}{2}, so \cos θ will be negative] = - (s e c x - \tan x) = - s e c x + \tan x$

Page No 5.40:

Question 4:

If π < x <2π, then $\sqrt{\frac{1 + \cos x}{1 - \cos x}}$ is equal to
(a) cosec x + cot x
(b) cosec x − cot x
(c) −cosec x + cot x
(d) −cosec x − cot x

Answer:

(d) −cosec x − cot x

$\sqrt{\frac{1 + co s x}{1 - co s x}} = \sqrt{\frac{(1 + co s x) (1 + co s x)}{(1 - co s x) (1 + co s x)}} = \sqrt{\frac{{(1 + co s x)}^{2}}{1 - co s^{2} x}} = \sqrt{\frac{{(1 + co s x)}^{2}}{\sin^{2} x}} = \frac{(1 + co s x)}{- \sin x} [as, π < x < 2 π, so \sin x will be negative] = - (c o s e c x + c o t x) = - c o s e c x - c o t x$

Page No 5.41:

Question 5:

If $0 < x < \frac{π}{2}$ , and if $\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}$ , then y is equal to
(a) $\cot \frac{x}{2}$

(b) $\tan \frac{x}{2}$

(c) $\cot \frac{x}{2} + \tan \frac{x}{2}$

(d) $\cot \frac{x}{2} - \tan \frac{x}{2}$

Answer:

(b) $\tan \frac{x}{2}$

$We have : \frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \Rightarrow \frac{y + 1}{1 - y} = \sqrt{\frac{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}} \Rightarrow \frac{y + 1}{1 - y} = \sqrt{\frac{{(co s \frac{x}{2} + \sin \frac{x}{2})}^{2}}{{(co s \frac{x}{2} - \sin \frac{x}{2})}^{2}}} \Rightarrow \frac{y + 1}{1 - y} = \frac{(co s \frac{x}{2} + \sin \frac{x}{2})}{(co s \frac{x}{2} - \sin \frac{x}{2})} [∵ 0 < x < \frac{π}{2} \Rightarrow 0 < \frac{x}{2} < \frac{π}{4}, 0 to \frac{π}{4} \cos x is greater than \sin x] \Rightarrow \frac{y + 1}{1 - y} = \frac{\frac{co s \frac{x}{2}}{co s \frac{x}{2}} + \frac{\sin \frac{x}{2}}{co s \frac{x}{2}}}{\frac{co s \frac{x}{2}}{co s \frac{x}{2}} - \frac{\sin \frac{x}{2}}{co s \frac{x}{2}}} \Rightarrow \frac{1 + y}{1 - y} = \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} Comparing both the sides : y = \tan \frac{x}{2}$

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Question 6:

If $\frac{π}{2} < x < π, then \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}$ is equal to
(a) 2 sec x
(b) −2 sec x
(c) sec x
(d) −sec x

Answer:

(b) −2 sec x

$\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = \sqrt{\frac{(1 - \sin x) (1 - \sin x)}{(1 + \sin x) (1 - \sin x)}} + \sqrt{\frac{(1 + \sin x) (1 + \sin x)}{(1 - \sin x) (1 + \sin x)}} = \sqrt{\frac{{(1 - \sin x)}^{2}}{1 - \sin^{2} x}} + \sqrt{\frac{{(1 + \sin x)}^{2}}{1 - \sin^{2} x}} = \sqrt{\frac{{(1 - \sin x)}^{2}}{\cos^{2} x}} + \sqrt{\frac{{(1 + \sin x)}^{2}}{\cos^{2} x}} = \frac{(1 - \sin x)}{- co s x} + \frac{(1 + \sin x)}{- co s x} [\frac{π}{2} < x < π, so \cos x will be negative .] = - (s e c x - \tan x) - (s e c x + \tan x) = - 2 s e c x$

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Question 7:

If x = r sin θ cos Ï•, y = r sin θ sin Ï• and z = r cos θ, then x² + y² + z² is independent of
(a) θ, Ï•
(b) r, θ
(c) r, Ï•
(d) r.

Answer:

(a) θ, Ï•

We have:
x = r sin θ cos Ï• , y = r sin θ sin Ï• and z = r cos θ,
∴ x² + y² + z²
$= {(r \sin θ \cos ϕ)}^{2} + {(r \sin θ \sin ϕ)}^{2} + {(r \cos θ)}^{2} = r^{2} \sin^{2} θ \cos^{2} ϕ + r^{2} \sin^{2} θ \sin^{2} ϕ + r^{2} \cos^{2} θ = r^{2} \sin^{2} θ (\cos^{2} ϕ + \sin^{2} ϕ) + r^{2} \cos^{2} θ = r^{2} \sin^{2} θ \times 1 + r^{2} \cos^{2} θ = r^{2} \sin^{2} θ + r^{2} \cos^{2} θ = r^{2} (\sin^{2} θ + \cos^{2} θ) = r^{2} \times 1 = r^{2} Thus, x^{2} + y^{2} + z^{2} is independent of θ and ϕ .$

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Question 8:

If tan x + sec x = $\sqrt{3}$ , 0 < x < π, then x is equal to

(a) $\frac{5 π}{6}$

(b) $\frac{2 π}{3}$

(c) $\frac{π}{6}$

(d) $\frac{π}{3}$

Answer:

(c) $\frac{π}{6}$

$We have : \tan x + s e c x = \sqrt{3} [0 < x < π] \Rightarrow s e c x + \tan x = \sqrt{3} \Rightarrow \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sqrt{3} \Rightarrow 1 + \sin x = \sqrt{3} \cos x$
$\Rightarrow {(1 + \sin x)}^{2} = {(\sqrt{3} \cos x)}^{2} \Rightarrow 1 + \sin^{2} x + 2 \sin x = 3 \cos^{2} x \Rightarrow 1 + \sin^{2} x + 2 \sin x = 3 (1 - \sin^{2} x) \Rightarrow 4 \sin^{2} x + 2 \sin x = 2 \Rightarrow 2 \sin^{2} x + \sin x - 1 = 0 \Rightarrow \sin x = - 1, \frac{1}{2} Since 0 < x < π, \sin x cannot be negative . ∴ \sin x = \frac{1}{2} ∴ x = \frac{π}{6}$

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Question 9:

If tan $x = - \frac{1}{\sqrt{5}}$ and θ lies in the IV quadrant, then the value of cos x is
(a) $\frac{\sqrt{5}}{\sqrt{6}}$

(b) $\frac{2}{\sqrt{6}}$

(c) $\frac{1}{2}$

(d) $\frac{1}{\sqrt{6}}$

Answer:

(a) $\frac{\sqrt{5}}{\sqrt{6}}$

$In the fourth quadrant, \cos x and \sec x are positive . \cos x = \frac{1}{s e c x} = \frac{1}{\sqrt{s e c^{2} x}} = \frac{1}{\sqrt{1 + \tan^{2} x}} = \frac{1}{\sqrt{1 + {(- \frac{1}{\sqrt{5}})}^{2}}} = \frac{1}{\sqrt{\frac{6}{5}}} = \frac{\sqrt{5}}{\sqrt{6}}$

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Question 10:

If $\frac{3 π}{4} < α < π, then \sqrt{2 \cot α + \frac{1}{\sin^{2} α}}$ is equal to
(a) 1 − cot α
(b) 1 + cot α
(c) −1 + cot α
(d) −1 −cot α

Answer:

(d) −1 −cot α

$We have : \sqrt{2 cotα + \frac{1}{\sin^{2} α}} = \sqrt{\frac{2 \cos α}{\sin α} + \frac{1}{\sin^{2} α}} = \sqrt{\frac{2 \sin α \cos α + 1}{\sin^{2} α}} = \sqrt{\frac{2 \sin α \cos α + \sin^{2} α + \cos^{2} α}{\sin^{2} α}} = \sqrt{\frac{{(\sin α + \cos α)}^{2}}{\sin^{2} α}} = \sqrt{{(1 + \cot α)}^{2}} = |1 + \cot α| = - (1 + \cot α) [When \frac{3 π}{4} < α < π, \cot α < - 1 \Rightarrow c o t α + 1 < 0] = - 1 - c o t α$

Page No 5.41:

Question 11:

sin⁶A + cos⁶A + 3 sin²A cos²A =
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(b) 1

$We have : \sin^{6} A + \cos^{6} A + 3 (\sin^{2} A) (\cos^{2} A) = {(\sin^{2} A)}^{3} + {(\cos^{2} A)}^{3} + 3 (\sin^{2} A) (\cos^{2} A) \times 1 = {(\sin^{2} A)}^{3} + {(\cos^{2} A)}^{3} + 3 (\sin^{2} A) (\cos^{2} A) (\sin^{2} A + \cos^{2} A) = {(\sin^{2} A + \cos^{2} A)}^{3} = 1^{3} = 1$

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Question 12:

If $cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{π}{2},$ , then cos x is equal to
(a) $\frac{5}{3}$

(b) $\frac{3}{5}$

(c) $- \frac{3}{5}$

(d) $- \frac{5}{3}$

Answer:

(b) $\frac{3}{5}$

$We have : cosec x - \cot x = \frac{1}{2} (1) \Rightarrow \frac{1}{cosec x - \cot x} = 2 \Rightarrow \frac{{cosec}^{2} x - \cot^{2} x}{cosec x - \cot x} = 2 \Rightarrow \frac{(cosec x + \cot x) (cosec x - \cot x)}{(cosec x - \cot x)} = 2 ∴ \cos e c x + \cot x = 2 (2) Adding (1) and (2) : 2 \cos ec x = \frac{1}{2} + 2 \Rightarrow 2 cosec x = \frac{5}{2} \Rightarrow cosec x = \frac{5}{4} \Rightarrow \frac{1}{\sin x} = \frac{5}{4} \Rightarrow \sin x = \frac{4}{5}$
$Now, 0 < θ < \frac{π}{2} ∴ \cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - {(\frac{4}{5})}^{2}} = \frac{3}{5}$

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Question 13:

If $cosec x + \cot x = \frac{11}{2}$ , then tan x =
(a) $\frac{21}{22}$

(b) $\frac{15}{16}$

(c) $\frac{44}{117}$

(d) $\frac{117}{44}$

Answer:

(c) $\frac{44}{117}$

$We have : cosec x + cot x = \frac{11}{2} (1) \Rightarrow \frac{1}{cosec x + cot x} = \frac{2}{11} \Rightarrow \frac{{cosec}^{2} x - {cot}^{2} x}{cosec x + cot x} = \frac{2}{11} \Rightarrow \frac{(cosec x + cot x) (cosec x - cot x)}{(cosec x + cot x)} = \frac{2}{11} ∴ \cos ec A - \cot x = \frac{2}{11} (2) Subtracting (2) from (1) : 2 cot x = \frac{11}{2} - \frac{2}{11} \Rightarrow 2 cot x = \frac{121 - 4}{22} \Rightarrow 2 cot x = \frac{117}{22} \Rightarrow \cot x = \frac{117}{44} \Rightarrow \frac{1}{\tan x} = \frac{117}{44} \Rightarrow \tan x = \frac{44}{117}$

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Question 14:

$\sec^{2} x = \frac{4 x y}{(x + y)^{2}}$ is true if and only if
(a) x + y ≠ 0
(b) x = y, x ≠ 0
(c) x = y
(d) x ≠0, y ≠ 0

Answer:

(b) x = y, x ≠ 0

$We have : \sec^{2} x = \frac{4 x y}{(x + y)^{2}} \Rightarrow \frac{4 x y}{(x + y)^{2}} \geq 1 [∵ \sec^{2} x \geq 1] \Rightarrow 4 x y \geq (x + y)^{2}$
$\Rightarrow 4 x y \geq x^{2} + y^{2} + 2 x y \Rightarrow 2 x y \geq x^{2} + y^{2} \Rightarrow {(x - y)}^{2} \leq 0 \Rightarrow (x - y) \leq 0 \Rightarrow x = y For x = 0, s e c^{2} x will not be defined, \Rightarrow x \neq 0 ∴ x = y$

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Question 15:

If x is an acute angle and $\tan x = \frac{1}{\sqrt{7}}$ , then the value of $\frac{{cosec}^{2} x - \sec^{2} x}{{cosec}^{2} x + \sec^{2} x}$ is
(a) 3/4
(b) 1/2
(c) 2
(d) 5/4

Answer:

(a) 3/4

$We have : \tan x = \frac{1}{\sqrt{7}} ∴ \tan^{2} x = \frac{1}{7} Now, dividing the numerator and the denominator of \frac{{cosec}^{2} x - \sec^{2} x}{{cosec}^{2} x + \sec^{2} x} by {cosec}^{2} x :$
$\frac{1 - \tan^{2} x}{1 + \tan^{2} x} = \frac{1 - \frac{1}{7}}{1 + \frac{1}{7}} = \frac{6}{8} = \frac{3}{4}$

Page No 5.41:

Question 16:

The value of sin²5° + sin²10° + sin²15° + ... + sin²85° + sin²90° is
(a) 7
(b) 8
(c) 9.5
(d) 10

Answer:

(c) 9.5

$We have : \sin^{2} 5 ° + \sin^{2} 10 ° + \sin^{2} 15 ° + . . . + \sin^{2} 85 ° + \sin^{2} 90 ° = \sin^{2} 5 ° + \sin^{2} 10 ° + \sin^{2} 15 ° + . . . + \sin^{2} (90 ° - 10 °) + \sin^{2} (90 ° - 5 °) + \sin^{2} 90 ° = \sin^{2} 5 ° + \sin^{2} 10 ° + \sin^{2} 15 ° + . . . + \cos^{2} 10 ° + \cos^{2} 5 ° + \sin^{2} 90 ° = (\sin^{2} 5 ° + \cos^{2} 5 °) + (\sin^{2} 10 ° + \cos^{2} 10 °) + + (\sin^{2} 15 ° + \cos^{2} 15 °) + (\sin^{2} 20 ° + \cos^{2} 20 °) + (\sin^{2} 25 ° + \cos^{2} 25 °) + (\sin^{2} 30 ° + \cos^{2} 30 °) + (\sin^{2} 35 ° + \cos^{2} 35 °) + (\sin^{2} 40 ° + \cos^{2} 40 °) + \sin^{2} 45 ° + \sin^{2} 90 ° = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + {(\frac{1}{\sqrt{2}})}^{2} + {(1)}^{2} [∵ \sin^{2} θ + \cos^{2} θ = 1] = 8 + \frac{1}{2} + 1 = 9.5$

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Question 17:

sin²π/18 + sin²π/9 + sin² 7π/18 + sin² 4π/9 =
(a) 1
(b) 4
(c) 2
(d) 0

Answer:

(c) 2

$We have : \sin^{2} \frac{π}{18} + \sin^{2} \frac{π}{9} + \sin^{2} \frac{7 π}{18} + \sin^{2} \frac{4 π}{9} = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \sin^{2} \frac{7 π}{18} + \sin^{2} \frac{8 π}{18} = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \sin^{2} (\frac{7 π}{18}) + \sin^{2} (\frac{8 π}{18}) = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \sin^{2} (\frac{π}{2} - \frac{2 π}{18}) + \sin^{2} (\frac{π}{2} - \frac{π}{18}) = \sin^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \cos^{2} \frac{2 π}{18} + \cos^{2} \frac{π}{18} = \sin^{2} \frac{π}{18} + \cos^{2} \frac{π}{18} + \sin^{2} \frac{2 π}{18} + \cos^{2} \frac{2 π}{18} = 1 + 1 = 2$

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Question 18:

If tan A + cot A = 4, then tan⁴A + cot⁴A is equal to
(a) 110
(b) 191
(c) 80
(d) 194

Answer:

(d) 194

$We have : \tan A + \cot A = 4 Squaring both the sides : {(\tan A + \cot A)}^{2} = 4^{2} \Rightarrow \tan^{2} A + \cot^{2} A + 2 (\tan A) (\cot A) = 16 \Rightarrow \tan^{2} A + \cot^{2} A + 2 = 16 \Rightarrow \tan^{2} A + \cot^{2} A = 14 Squaring both the sides again : {(\tan^{2} A + \cot^{2} A)}^{2} = 14^{2} \Rightarrow \tan^{4} A + \cot^{4} A + 2 (\tan^{2} A) (\cot^{2} A) = 196 \Rightarrow \tan^{4} A + \cot^{4} A + 2 = 196 \Rightarrow \tan^{4} A + \cot^{4} A = 194$

Page No 5.42:

Question 19:

If x sin 45° cos² 60° = $\frac{\tan^{2} 60 ° cosec 30 °}{\sec 45 ° \cot^{2 °} 30 °}$ , then x =
(a) 2
(b) 4
(c) 8
(d) 16

Answer:

(c) 8

$We have : x \sin 45 ° \cos^{2} 60 ° = \frac{\tan^{2} 60 ° \cos e c 30 °}{\sec 45 ° \cot^{2} 30 °} \Rightarrow x \times (\frac{1}{\sqrt{2}}) \times {(\frac{1}{2})}^{2} = \frac{{(\sqrt{3})}^{2} \times (2)}{(\sqrt{2}) \times {(\sqrt{3})}^{2}} \Rightarrow \frac{x}{4 \sqrt{2}} = \frac{6}{3 \sqrt{2}} \Rightarrow x = \frac{6}{3 \sqrt{2}} \times 4 \sqrt{2} \Rightarrow x = 8$

Page No 5.42:

Question 20:

If A lies in second quadrant 3tanA + 4 = 0, then the value of 2cotA − 5cosA + sinA is equal to

(a) $- \frac{53}{10}$ (b) $\frac{23}{10}$ (c) $\frac{37}{10}$ (d) $\frac{7}{10}$

Answer:

It is given that $\frac{π}{2} < A < π$ .

$3 \tan A + 4 = 0 \Rightarrow \tan A = - \frac{4}{3} \Rightarrow \cot A = - \frac{3}{4}$

Now,

$\sec A = \pm \sqrt{1 + \tan^{2} A} = \pm \sqrt{1 + \frac{16}{9}} = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3} ∴ \sec A = - \frac{5}{3} (A lies in 2 nd quadrant) \Rightarrow \cos A = - \frac{3}{5}$

Also,

$\sin A = \pm \sqrt{1 - \cos^{2} A} = \pm \sqrt{1 - \frac{9}{25}} = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} ∴ \sin A = \frac{4}{5} (A lies in 2 nd quadrant)$

So,

$2 \cot A - 5 \cos A + \sin A = 2 \times (- \frac{3}{4}) - 5 \times (- \frac{3}{5}) + \frac{4}{5} = - \frac{3}{2} + 3 + \frac{4}{5} = \frac{- 15 + 30 + 8}{10} = \frac{23}{10}$

Hence, the correct answer is option B.

Page No 5.42:

Question 21:

If $\cos e c x + c o t x = \frac{11}{2}$ , then tan x =
(a) $\frac{21}{22}$

(b) $\frac{15}{16}$

(c) $\frac{44}{117}$

(d) $\frac{117}{43}$

Answer:

(c) $\frac{44}{117}$

$We have : cosec x + \cot x = \frac{11}{2} (1) \Rightarrow \frac{1}{cosec x + \cot x} = \frac{2}{11} \Rightarrow \frac{{cosec}^{2} x - \cot^{2} x}{cosec x + \cot x} = \frac{2}{11} \Rightarrow \frac{(cosec x + \cot x) (cosec x - \cot x)}{(cosec x + \cot x)} = \frac{2}{11} ∴ c o s e c x - c o t x = \frac{2}{11} (2) Subtracting (2) from (1) : 2 c o t x = \frac{11}{2} - \frac{2}{11} \Rightarrow 2 c o t x = \frac{121 - 4}{22} \Rightarrow 2 c o t x = \frac{117}{22} \Rightarrow c o t x = \frac{117}{44} \Rightarrow \frac{1}{\tan x} = \frac{117}{44} \Rightarrow \tan x = \frac{44}{117}$

Page No 5.42:

Question 22:

If tan θ + sec θ =e^x, then cos θ equals

(a) $\frac{e^{x} + e^{- x}}{2}$
(b) $\frac{2}{e^{x} + e^{- x}}$
(c) $\frac{e^{x} - e^{- x}}{2}$
(d) $\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

Answer:

(b) $\frac{2}{e^{x} + e^{- x}}$

$We have : \tan θ + \sec θ = e^{x} \sec θ + \tan θ = e^{x} (1) \Rightarrow \frac{1}{secθ + tanθ} = \frac{1}{e^{x}} \Rightarrow \frac{\sec^{2} θ - \tan^{2} θ}{\sec θ + \tan θ} = \frac{1}{e^{x}} \Rightarrow \frac{(\sec θ + \tan θ) (\sec θ - \tan θ)}{(\sec θ + \tan θ)} = \frac{1}{e^{x}} ∴ s e c θ - \tan θ = \frac{1}{e^{x}} (2) Adding (1) and (2) : 2 \sec θ = e^{x} + \frac{1}{e^{x}} \Rightarrow 2 \sec θ = \frac{{(e^{x})}^{2} + 1}{e^{x}} \Rightarrow \sec θ = \frac{e^{2 x} + 1}{2 e^{x}} \Rightarrow \sec θ = \frac{1}{2} \times \frac{e^{2 x} + 1}{e^{x}} \Rightarrow \sec θ = \frac{1}{2} \times (e^{x} + e^{- x}) \Rightarrow \frac{1}{\cos θ} = \frac{e^{x} + e^{- x}}{2} \Rightarrow \cos θ = \frac{2}{e^{x} + e^{- x}}$

Page No 5.42:

Question 23:

If sec x + tan x = k, cos x =

(a) $\frac{k^{2} + 1}{2 k}$

(b) $\frac{2 k}{k^{2} + 1}$

(c) $\frac{k}{k^{2} + 1}$

(d) $\frac{k}{k^{2} - 1}$

Answer:

(b) $\frac{2 k}{k^{2} + 1}$

$We have : \sec x + \tan x = k (1) \Rightarrow \frac{1}{\sec x + \tan x} = \frac{1}{k} \Rightarrow \frac{\sec^{2} x - \tan^{2} x}{\sec x + \tan x} = \frac{1}{k} \Rightarrow \frac{(\sec x + \tan x) (\sec x - \tan x)}{(\sec x + \tan x)} = \frac{1}{k} ∴ s e c x - \tan x = \frac{1}{k} (2) Adding (1) and (2) : 2 \sec x = k + \frac{1}{k} \Rightarrow 2 \sec x = \frac{k^{2} + 1}{k} \Rightarrow \sec x = \frac{k^{2} + 1}{2 k} \Rightarrow \frac{1}{\cos x} = \frac{k^{2} + 1}{2 k} \Rightarrow \cos x = \frac{2 k}{k^{2} + 1}$

Page No 5.42:

Question 24:

If $f (x) = \cos^{2} x + \sec^{2} x$ , then

(a) f(x) < 1 (b) f(x) = 1 (c) 1 < f(x) < 2 (d) f(x) ≥ 2

Answer:

$f (x) = \cos^{2} x + \sec^{2} x = \cos^{2} x + \sec^{2} x - 2 \cos x \sec x + 2 \cos x \sec x = {(\sec x - \cos x)}^{2} + 2 ∴ f (x) \geq 2 \forall x [{(\sec x - \cos x)}^{2} \geq 0 \forall x]$

Hence, the correct option is answer D.

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Question 25:

Which of the following is incorrect?

(a) $\sin x = - \frac{1}{5}$ (b) cos x = 1 (c) $\sec x = \frac{1}{2}$ (d) tan x = 20

Answer:

(a) $\sin x = - \frac{1}{5}$ is correct as $- 1 \leq \sin x \leq 1$

(b) cos x = 1 is correct as $- 1 \leq \cos x \leq 1$

(c) $\sec x = \frac{1}{2}$ is not correct as $\sec x \in (- \infty, - 1] \cup [1, \infty)$

(d) tan x = 20 is correct as tan x can take any real value.

Hence, the correct answer is option C.

Page No 5.42:

Question 26:

The value of $\cos 1 ° \cos 2 ° \cos 3 ° . . . \cos 179 °$ is

(a) $\frac{1}{\sqrt{2}}$ (b) 0 (c) 1 (d) $- 1$

Answer:

$\cos 1 ° \cos 2 ° \cos 3 ° . . . \cos 179 ° = \cos 1 ° \cos 2 ° \cos 3 ° . . . \cos 90 ° . . . \cos 179 ° = 0 (\cos 90 ° = 0)$

Hence, the correct answer is option B.

Page No 5.42:

Question 27:

The value of $\tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 89 °$ is

(a) 0 (b) 1 (c) $\frac{1}{2}$ (d) not defined

Answer:

We know that, $\tan (90 ° - θ) = \cot θ$

So,

$\tan 89 ° = \tan (90 ° - 1 °) = \cot 1 ° \tan 88 ° = \tan (90 ° - 2 °) = \cot 2 ° \tan 87 ° = \tan (90 ° - 3 °) = \cot 3 ° . . . . . . . . \tan 46 ° = \tan (90 ° - 44 °) = \cot 44 °$

$∴ \tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 89 ° = \tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 44 ° \tan 45 ° \tan 46 ° . . . \tan 87 ° \tan 88 ° \tan 89 ° = \tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 44 ° \tan 45 ° \cot 44 ° . . . \cot 3 ° \cot 2 ° \cot 1 ° = (\tan 1 ° \cot 1 °) (\tan 2 ° \cot 2 °) (\tan 3 ° \cot 3 °) . . . (\tan 44 ° \cot 44 °) \tan 45 ° = 1 (\tan 45 ° = 1 and \tan θ \cot θ = 1)$

Hence, the correct answer is option B.

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Question 28:

Which of the following is correct?

(a) $\sin 1 ° > \sin 1$ (b) $\sin 1 ° < \sin 1$ (c) $\sin 1 ° = \sin 1$ (d) $\sin 1 ° = \frac{π}{180} \sin 1$

Answer:

We know that, 1 radian is approximately 57º.

Also, the value of sinx is always increasing for $0 \leq x \leq 90 °$ ( or sinx is an increasing function for $0 \leq x \leq 90 °$ ).

Now,

$1 ° < 57 ° Or 1 ° < 1 radian ∴ \sin 1 ° < \sin 1$

Hence, the correct answer is option B.

Page No 5.42:

Question 29:

If $\tan θ = - \frac{4}{3},$ then sin θ is equal to

(a) $- \frac{4}{5}$ but not $\frac{4}{5}$

(b) $- \frac{4}{5}$ or $\frac{4}{5}$

(c) $\frac{4}{5}$ but not $- \frac{4}{5}$

(d) none of these

Answer:

Given $\tan θ = - \frac{4}{3}$
Since tan is negative in 2nd or 4th quadrant
⇒ θ lies in 2nd or 4th quadrant

Since AC² = AB² + BC² = (–4)² + (3)²
AC² = 25
AC = ± 5
i.e. AC = 5
$∴ sinθ = - \frac{4}{5} or \frac{4}{5}$
Hence, the correct answer is option B.

Page No 5.42:

Question 30:

If sin θ and cos θ are the roots of the equation ax² – bx + c = 0, then a, b and c satisfy the relation
(a) a² + b² + 2ac = 0
(b) a² – b² + 2ac = 0
(c) a² + c² + 2ab = 0
(d) a² – b² – 2ac = 0

Answer:

Given sinθ and cosθ are roots of ax² – bx + c = 0
Sum of roots is $\frac{b}{a}$ and product of root is $\frac{c}{a}$
i.e $\cos θ + \sin θ = \frac{b}{a} and \cos θ \sin θ = \frac{c}{a}$
Since sin²θ + cos²θ = 1
$i . e {(\sin θ + \cos θ)}^{2} = \sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ i . e {(\frac{b}{a})}^{2} = 1 + 2 \frac{c}{a} i . e b^{2} = a^{2} + 2 a c i . e a^{2} - b^{2} + 2 a c$

Hence, the correct answer is option B.

Page No 5.43:

Question 31:

If sin θ + cosec θ = 2, then sin²θ + cosec²θ is equal to
(a) 1
(b) 4
(c) 2
(d) none of these

Answer:

Given sinθ + cosecθ = 2
⇒ (sinθ + cosecθ)² = 4
i.e sin²θ + cosec²θ + 2 sinθ cosecθ = 4
$i . e \sin^{2} θ + {cosec}^{2} θ + 2 \sin θ \frac{1}{\sin θ} = 4 i . e \sin^{2} θ + {cosec}^{2} θ = 2$

Hence, the correct answer is option C.

Page No 5.43:

Question 32:

Which of the following is incorrect?
(a) $\sin θ = - \frac{1}{5}$
(b) cosθ = 1
(c) $\sec θ = \frac{1}{2}$
(d) tanθ = 20

Answer:

$If \sec θ = \frac{1}{2} \Rightarrow \cos θ = 2 i . e θ = \cos^{- 1} (2)$
i.e No solution
$∴ \sec θ = \frac{1}{2} is incorrect$
Hence, the correct answer is option C.

Page No 5.43:

Question 33:

If for real values of $x, \cos θ = x + \frac{1}{x},$ then
(a) θ is an acute angle
(b) θ is a right angle
(c) θ is an obtuse angle
(d) No value of θ is possible

Answer:

Given for real value of x, $\cos θ = x + \frac{1}{x}$
$i . e \cos θ = \frac{x^{2} + 1}{x} \Rightarrow x \cos θ = x^{2} + 1 \Rightarrow x^{2} - x \cos θ + 1 = 0 for x \in ℝ i . e roots should be real i . e b^{2} - 4 a c \geq 0 i . e \cos^{2} θ - 4 (1) (1) \geq 0 i . e \cos^{2} θ \geq 4$
which is not possible (âˆµ –1 ≤ cosθ ≤ 1)
∴ No such value of θ is possible
Hence, the correct answer is option D.

Page No 5.43:

Question 1:

The value of $\frac{\sin 70 °}{\sin 110 °}$ is _____________.

Answer:

$\begin{array}{rcl} \frac{\sin 70 °}{\sin 110 °} & = & \frac{\sin 70 °}{\sin (90 ° + 20 °)} \\ = & \frac{\sin 70 °}{\cos 20 °} (Since \sin (90 ° + θ) = \cos θ) \\ = & \frac{\sin 70 °}{\cos (90 ° - 70 °)} \\ = & \frac{\sin 70 °}{\sin 70 °} = 1 (∵ \cos (90 ° - θ) = \sin θ) \end{array}$

$Hence \frac{\sin 70 °}{\sin 110 °} = 1$

Page No 5.43:

Question 2:

The value of $\frac{\cos 50 °}{\cos 130 °}$ is _____________.

Answer:

$\begin{array}{rcl} \frac{\cos 50 °}{\cos 130 °} & = & \frac{\cos 50 °}{\cos (180 ° - 50 °)} \\ = & \frac{\cos 50 °}{- \cos 50 °} (∵ \cos (180 ° - θ) = - \cos θ) \\ = & - 1 \end{array}$
$Hence \frac{\cos 50 °}{\cos 130 °} = - 1$

Page No 5.43:

Question 3:

The values of $f (x) = 2 \sin \sqrt{x^{2} + x + 1}$ lie in the interval __________ .

Answer:

$f (x) = 2 \sin \sqrt{x^{2} + x + 1} Since - 1 \leq \sin θ \leq 1 i . e - 1 \leq \sin \sqrt{x^{2} + x + 1} \leq 1 i . e - 2 \leq 2 \sin \sqrt{x^{2} + x + 1} \leq 2 i . e 2 \sin \sqrt{x^{2} + x + 1} \in [- 2, 2] i . e . 2 \sin \sqrt{x^{2} + x + 1} lies in interval [- 2, 2]$

Page No 5.43:

Question 4:

If sin x + cos x = a, then sin x – cos x = __________.

Answer:

Given sin x + cos x = a
i.e (sin x – cos x)²=a² (squaring both side)

$\Rightarrow \sin^{2} x + \cos^{2} x + 2 \sin x \cos x = a^{2} \Rightarrow 1 + 2 \sin x \cos x = a^{2} i . e 2 \sin x \cos x = a^{2} - 1$
Now, consider
$\begin{array}{rcl} {(\sin x - \cos x)}^{2} & = & \sin^{2} x + \cos^{2} x - 2 \sin x \cos x \\ = & 1 - (a^{2} - 1) \\ = & 1 - a^{2} + 1 \\ {(\sin x - \cos x)}^{2} & = & 2 - a^{2} \\ i . e \sin x - \cos x & = & \pm \sqrt{2 - a^{2}} \end{array}$

Page No 5.43:

Question 5:

If $- \frac{π}{2} < x < \frac{π}{2},$ then $\sqrt{\frac{1 - \sin x}{1 + \sin x}}$ is equal to _________.

Answer:

If $- \frac{π}{2} < x < \frac{π}{2}$ is given,
$\begin{array}{rcl} \sqrt{\frac{1 - \sin x}{1 + \sin x}} & = & \sqrt{\frac{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}} \\ (using identics : - \sin^{2} θ + \cos^{2} θ - 1 \sin 2 θ = 2 \sin θ \cos θ) \\ = & \sqrt{\frac{{(\sin \frac{x}{2} - \cos \frac{x}{2})}^{2}}{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}}} \\ = & \frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} \\ = & \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2} + 1} (dividing by \cos \frac{x}{2}) \\ = & \frac{\tan \frac{x}{2} - \tan \frac{π}{4}}{1 + \tan \frac{π}{4} \tan \frac{x}{2}} \\ = & \tan (\frac{x}{2} - \frac{π}{4}) \end{array}$
$i . e \sqrt{\frac{1 - \sin x}{1 + \sin x}} = \tan (\frac{x}{2} - \frac{π}{4})$

Page No 5.43:

Question 6:

If $\sec x + \tan x = \sqrt{3},$ then sec x – tan x = ___________.

Answer:

Given $\sec x + \tan x = \sqrt{3}$
$Since \sec^{2} x - \tan^{2} x = 1 (using identity) \Rightarrow (\sec x + \tan x) (\sec x - \tan x) = 1 \Rightarrow \sqrt{3} (\sec x - \tan x) = 1 \Rightarrow \sec x - \tan x = \frac{1}{\sqrt{3}}$

Page No 5.43:

Question 7:

If $\sec x - \tan x = \frac{2}{3},$ then tan x = ___________.

Answer:

$Given \sec x - \tan x = \frac{2}{3} (1) Since \sec^{2} x - \tan^{2} x = 1 \Rightarrow (\sec x - \tan x) (\sec x + \tan x) = 1 \Rightarrow \frac{2}{3} (\sec x + \tan x) = 1 \Rightarrow \sec x + \tan x = \frac{3}{2} (2)$
Subtracting (1) from (2)
$\sec x + \tan x - \sec x + \tan x = \frac{3}{2} - \frac{2}{3} \Rightarrow 2 \tan x = \frac{9 - 4}{6} \Rightarrow 2 \tan x = \frac{5}{6} \Rightarrow \tan x = \frac{5}{12}$

Page No 5.43:

Question 8:

If cosec x + cot x = α, then sin x = _________.

Answer:

Given cosec x + cot x = a (1)
Since cosec²x – cot²x = 1
(cosec x – cot x) (cosec x + cot x) = 1
i.e (cosec x – cot x) a = 1
$i . e cosec x - \cot x = \frac{1}{a} (2)$
adding (1) and (2)
$2 cosec x = a + \frac{1}{a} 2 cosec x = \frac{a^{2} + 1}{a} \frac{2 a}{a^{2} + 1} = \frac{1}{cosec x} i . e \sin x = \frac{2 a}{a^{2} + 1}$

Page No 5.43:

Question 9:

If $cosec x + \cot x = \frac{11}{2},$ then the value of tan x is ______________.

Answer:

$cosec x + \cot x = \frac{11}{2} (1) Since {cosec}^{2} x - \cot^{2} x = 1 \Rightarrow (cosec x - \cot x) (cosec x + \cot x) = 1 \Rightarrow cosec x - \cot x = \frac{2}{11} (2) Subtrating (2) from (1) cosec x + \tan x - cosec x + \cot x = \frac{11}{2} - \frac{2}{11} 2 \cot x = \frac{121 - 4}{22} = \frac{117}{22} \cot x = \frac{117}{44} i . e \tan x = \frac{44}{117}$

Page No 5.43:

Question 10:

If $\sin x = \frac{- 24}{25},$ then the value of tan x is __________ .

Answer:

Given $\sin x = \frac{- 24}{25}$ i.e x lies in III or IV quadrant

$\sin x = \frac{A B}{A C} = \frac{- 24}{25} Since A C^{2} = A B^{2} + B C^{2} i . e 25^{2} = {(24)}^{2} + B C^{2} \Rightarrow 625 = 576 + B C^{2} \Rightarrow B C^{2} = 49 B C = \pm 7 \Rightarrow \tan x = \frac{24}{7} or \tan x = \frac{- 24}{7}$

Page No 5.43:

Question 11:

If sin x + cosec x = 2, then sin²x + cosec²x = ___________ .

Answer:

Given sin x + cosec x = 2
Squaring both sides, (sin x + cosec x)² = 4
$i . e \sin^{2} x + {cosec}^{2} x + 2 \sin x cosec x = 4 i . e \sin^{2} x + {cosec}^{2} x + 2 \sin x \times \frac{1}{\sin x} = 4 i . e \sin^{2} x + {cosec}^{2} x + 2 = 4 i . e \sin^{2} x + {cosec}^{2} x = 2$

Page No 5.43:

Question 12:

The value of tan $x + \cot (π + x) + \cot (\frac{π}{2} + x) + \cot (2 π - x)$ is ______________ .

Answer:

$\tan x + \cot (π + x) + \cot (\frac{π}{2} + x) + \cot (2 π - x) = \tan x + (+ \cot x) + \cot (\frac{π}{2} + x) + \cot (2 π - x) Since \cot (π + x) = \frac{1}{\tan (π + x)} = \frac{1}{+ \tan x} = + \cot x and \cot (\frac{π}{2} + x) = \frac{1}{\tan (\frac{π}{2} + x)} = \frac{1}{- c o t x} = - \tan x and \cot (2 π - x) = \frac{1}{\tan (2 π - x)} = \frac{1}{- \tan x} = - \cot x = \tan x + \cot x - \tan x - \cot x = 0 Hence, \tan x + \cot (π + x) + \cot (\frac{π}{2} + x) + \cot (2 π - x) = 0$

Page No 5.43:

Question 13:

The value of 3 (sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶x + cos⁶x) is ___________.

Answer:

3 (sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶x + cos⁶x)
$= 3 {[{(\sin x - \cos x)}^{2}]}^{2} + 6 (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x) + 4 [{(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}] = 3 {[\sin^{2} x + \cos^{2} x - 2 \cos x \sin x]}^{2} + 6 (1 + 2 \sin x \cos x) + 4 [{(\sin^{2} x + \cos_{1}^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)] = 1 = 3 {[1 - 2 \cos x \sin x]}^{2} + 6 + 12 \sin x \cos x + 4 [1 - 3 \sin^{2} x \cos^{2} x]$
$= 3 [1 + 4 \cos^{2} x \sin^{2} x - 4 \cos x \sin x] + 6 + 12 \sin x \cos x + 4 - 12 \sin^{2} x \cos^{2} x = 3 + 12 \cos^{2} x \sin^{2} x - 12 \cos x \sin x + 6 + 12 \cos x \sin x + 4 - 12 \sin^{2} x \cos^{2} x = 13 Hence, 3 {(\sin x - \cos x)}^{4} + 6 {(\sin x + \cos x)}^{2} + 4 (\sin^{6} x + \cos^{6} x) has value 13 .$

Page No 5.43:

Question 14:

Given x > 0, the value of $f (x) = - 3 \cos \sqrt{3 + x + x^{2}}$ lie in the interval ____________.

Answer:

Given x > 0,
$f (x) = 3 \cos \sqrt{3 + x + x^{2}} Since - 1 \leq \cos θ \leq 1 for any θ i . e - 1 \leq \cos (\sqrt{3 + x + x^{2}}) \leq 1 i . e - 3 \leq 3 \cos (\sqrt{3 + x + x^{2}}) \leq 3 i . e 3 \cos (\sqrt{3 + x + x^{2}}) \in [- 3, 3]$

Page No 5.43:

Question 15:

If sin x + cos x = a, then sin⁶x + cos⁶x = __________.

Answer:

Given sin x + cos x = a
Squaring both sides, (sin x + cos x)² = a²
$\sin^{2} x + \cos^{2} x + 2 \sin x \cos x = a^{2} i . e 1 + 2 \sin x \cos x = a^{2} i . e 2 \sin x \cos x = a^{2} - 1 (1)$
Using identity, we have
$a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b) \sin^{6} x + \cos^{6} x = {(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3} = {(\sin^{2} x + \cos^{2} x)}^{3} - 3 \cos^{2} x \sin^{2} x (\sin^{2} x + \cos^{2} x) = {(1)}^{3} - 3 \cos^{2} x \sin^{2} x (1) = 1 - 3 {(\cos x \sin x)}^{2} = 1 - 3 {(\frac{a^{2} - 1}{2})}^{2} (from (1)) = 1 - \frac{3}{4} {(a^{2} - 1)}^{2} = 1 - \frac{3}{4} (a^{4} + 12 a^{2}) = 1 - [\frac{3}{4} a^{4} + \frac{3}{4} - \frac{3}{2} a^{2}] = 1 - \frac{3}{4} - \frac{3}{4} a^{4} + \frac{3}{2} a^{2} = \frac{1}{4} = \frac{3}{4} a^{4} + \frac{3}{2} a^{2}$
Hence, sin⁶x + cos⁶x

Page No 5.43:

Question 16:

The value of cos 1° cos 2° cos 3° _______cos 179° is ____________.

Answer:

cos 1° cos 2° cos 3° ....... cos 179°
= cos 1° cos 2° cos 3° ......... cos 90° cos 91° ......... cos 179°
= cos 1° × cos 2° × cos 3° ..... × 0 × cos 91° .......... cos 179° (âˆµ cos 90° = 0)
= 0 × finite
= 0
Hence, cos 1° cos 2° cos 3° ......... cos 179° = 0

Page No 5.43:

Question 17:

The value of tan 1° tan 2° tan 3° _________ tan 89° is _________.

Answer:

tan 1° tan 2° tan 3° ........ tan 89°
Since 89° = 90° – 1
88° = 90° – 2 ........... 46° = 90° – 44°
and tan 89° = tan (90° – 1°) = cot 1°
tan 88° = tan (90° – 2°) = cot 2°
.
.
.
tan 46° = tan (90° – 44°) = cot 44°
tan 45° = 1
∴ tan 1° tan 2° tan 3° .......... tan 45° cot 44° .....cot 3° cot 2° cot 1°
$= \tan 1 ° \tan 2 ° \tan 3 ° . . . . . \frac{1}{\tan 4.1 °} . . . . . \frac{1}{\tan 1 °}$
= 1
Hence, tan 1° tan 2° ....... tan 89° = 1.

Page No 5.44:

Question 18:

If $\frac{π}{2} < x < π and \sqrt{\frac{1 + \sin x}{1 - \sin x}} = k \sec x,$ then k = ___________.

Answer:

If $\frac{π}{2} < x < π$
Given $\sqrt{\frac{1 + \sin x}{1 - \sin x}} = k \sec x$
$\begin{array}{rcl} L . H . S \sqrt{\frac{1 + \sin x}{1 - \sin x}} & = & \sqrt{\frac{1 + \sin x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x}} (By multiplying dividing by 1 + \sin x) \\ = & \sqrt{\frac{(1 + \sin x)}{1 - \sin^{2} x}} \end{array}$
(correction)

Page No 5.44:

Question 19:

If $\frac{π}{2} < x < π and \sqrt{\frac{1 + \sin x}{1 - \sin x}} + \sqrt{\frac{1 - \sin x}{1 + \sin x}} = k \sec x,$ then k = ___________.

Answer:

$If \frac{π}{2} < x < π \sqrt{\frac{1 + \sin x}{1 - \sin x}} + \sqrt{\frac{1 - \sin x}{1 + \sin x}} = k \sec x (given)$
$\begin{array}{rcl} L . H . S & = & \sqrt{\frac{1 + \sin x}{1 - \sin x}} + \sqrt{\frac{1 - \sin x}{1 + \sin x}} \\ = & \sqrt{\frac{1 + \sin x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x}} + \sqrt{\frac{1 - \sin x}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x}} \\ = & \sqrt{[\frac{{(1 + \sin x)}^{2}}{1 - \sin^{2} x}]} + \sqrt{\frac{{(1 - \sin x)}^{2}}{1 - \sin^{2} x}} \\ = & \sqrt{\frac{{(1 + \sin x)}^{2}}{\cos^{2} x}} + \sqrt{\frac{{(1 - \sin x)}^{2}}{\cos^{2} x}} \\ = & \frac{1 + \sin x}{\sqrt{\cos^{2} x}} + \frac{(1 - \sin x)}{\sqrt{\cos^{2} x}} \\ = & \frac{2}{\sqrt{\cos^{2} x}} \end{array}$
$Since \frac{π}{2} < x < π, | \cos x | = - \cos x = \frac{2}{- \cos x} = - 2 \sec x = R . H . S (given) = k \sec x \Rightarrow Value of k = - 2$

Page No 5.44:

Question 20:

If $π < x < 2 π and \sqrt{\frac{1 + \cos x}{1 - \cos x}} + \sqrt{\frac{1 - \cos x}{1 + \cos x}} = k cosec x,$ then k = ___________.

Answer:

If π < x < 2π
$Given \sqrt{\frac{1 + \cos x}{1 - \cos x}} + \sqrt{\frac{1 - \cos x}{1 + \cos x}} = k cosec x$
$L . H . S is \sqrt{\frac{1 + \cos x}{1 - \cos x}} + \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{1 + \cos x}{1 - \cos x} \times \frac{1 + \cos x}{1 + \cos x}} + \sqrt{\frac{1 - \cos x}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x}} = \sqrt{\frac{{(1 + \cos x)}^{2}}{1 - \cos^{2} x}} + \sqrt{\frac{{(1 - \cos x)}^{2}}{1 - \cos^{2} x}} = \frac{(1 + \cos x)}{\sqrt{\sin^{2} x}} + \frac{(1 - \cos x)}{\sqrt{\sin^{2} x}} = \frac{2}{\sqrt{\sin^{2} x}} Since π < x < 2 π | \sin x | = - \sin x = \frac{2}{- \sin x} = - 2 cosec x = R . H . S (given) = k cosec x i . e k = - 2$

Page No 5.44:

Question 21:

The minimum value of 9 tan²θ + 4 cot²θ is ____________.

Answer:

9 tan²θ + 4 cot²θ
Since Arithmetic mass ≥ Geometric mean for 2 tans.
$i . e \frac{a + b}{2} \geq \sqrt{a b} Let a = 9 \tan^{2} θ b = 4 \cot^{2} θ \Rightarrow \frac{9 \tan^{2} θ + 4 \cot^{2} θ}{2} \geq \sqrt{9 \tan^{2} θ \times 4 \cot^{2} θ} = \sqrt{9 \times 4} = 3 \times 2 i . e 9 \tan^{2} θ + 4 \cot^{2} θ \geq 2 \times 6 = 12$

i.e minimum value of 9 tan²θ + 4 cot²θ is 12.

Page No 5.44:

Question 22:

If sec x = m and tan x = n, then $\frac{1}{m} \{(m + n) + \frac{1}{m + n}\}$ is equal to ____________.

Answer:

If sec x = m and tan x = n
$\frac{1}{m} \{(m + n) + \frac{1}{m + n}\} i . e \frac{1}{\sec x} \{\sec x + \tan x + \frac{1}{\sec x + \tan x}\} = \frac{1}{\sec x} \{\sec x + \tan x + \frac{1}{\sec x + \tan x} \times (\frac{\sec x - \tan x}{\sec x - \tan x})\} = \frac{1}{\sec x} \{\sec x + \tan x + \frac{\sec x - \tan x}{\sec^{2} x - \tan^{2} x}\} = \frac{1}{\sec x} \{\sec x + \tan x + \frac{\sec x - \tan x}{1}\} = \frac{1}{\sec x} \{2 \sec x\} = 2 ∴ \frac{1}{m} \{(m + n) + \frac{1}{m + n}\} = 2$

Page No 5.44:

Question 23:

If cos²x + sin x + 1 = 0, and 0 < x < 2π then x = _________.

Answer:

Given,
$\cos^{2} x + \sin x + 1 = 0 and 0 < x < 2 π 1 - \sin^{2} x + \sin x + 1 = 0 i . e (1 - \sin x) (1 + \sin x) + (1 + \sin x) = 0 (1 + \sin x) [1 - \sin x + 1] = 0 (1 + \sin x) [2 - \sin x] = 0 i . e \sin x = - 1 or \sin x = 2 since \sin x \neq 2 \Rightarrow \sin x = - 1 (only possibility) i . e x = \frac{3 π}{2}$

Page No 5.44:

Question 24:

If $\sin x = \frac{2 t}{1 + t^{2}}$ and x lies in the second quadrant, then cos x = ___________.

Answer:

Given $\sin x = \frac{2 t}{1 + t^{2}}$ and x lies in 2nd quadrant
Since sin²x + cos²x = 1
cos²x = 1 – sin²x
Since x lies in II Quadrant
⇒ cos x < 0
$∴ \cos x = - \sqrt{1 - \sin^{2} θ} = - \sqrt{1 - {(\frac{2 t}{1 + t^{2}})}^{2}} = - \sqrt{1 - \frac{4 t^{2}}{{(1 + t^{2})}^{2}}} = - \sqrt{\frac{{(1 + t^{2})}^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}} = - \sqrt{\frac{1 + t^{4} + 2 t^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}} = - \sqrt{\frac{1 + t^{4} - 2 t^{2}}{{(1 + t^{2})}^{2}}} = - \sqrt{\frac{{(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}}} Hence, \cos x = - \frac{(1 - t^{2})}{1 + t^{2}}$

Page No 5.44:

Question 25:

If $\sec x = t + \frac{1}{4 t},$ then the value of sec x + tan x is _________.

Answer:

$Given \sec x = t + \frac{1}{4 t} Since \tan^{2} x = \sec^{2} x - 1 = {(t + \frac{1}{4 t})}^{2} - 1 = t^{2} + \frac{1}{16 t^{2}} + \frac{1}{2} - 1 = t^{2} + \frac{1}{16 t^{2}} - \frac{1}{2} \tan^{2} x = {(t - \frac{1}{4 t})}^{2} i . e \tan x = \pm (t - \frac{1}{4 t}) ∴ \sec x + \tan x = (t + \frac{1}{4 t}) \pm (t - \frac{1}{4 t}) Hence, \sec x + \tan x = 2 t or \frac{1}{2 t}$

Page No 5.44:

Question 26:

If tan x + cot x = 4, then tan⁴x + cot⁴x = ___________.

Answer:

Given tan x + cot x = 4
$Since {(\tan^{2} x + \cot^{2} x)}^{2} = \tan^{4} x + \cot^{4} x + 2 \cot^{2} x \tan^{2} x i . e {(\tan^{2} x + \cot^{2} x)}^{2} = \tan^{4} x + \cot^{4} x + 2 i . e \tan^{4} x + \cot^{4} x = {(\tan^{2} x + \cot^{2} x)}^{2} - 2 (1) also, {(\tan x + \cot x)}^{2} = {(4)}^{2} i . e \tan^{2} x + \cot^{2} x + 2 \cot x \tan x = 16$
$\tan^{2} x + \cot^{2} x + 2 = 16 \tan^{2} x + \cot^{2} x = 14 ∴ equation (1), reduces to \tan^{4} x + \cot^{4} x = {(14)}^{2} - 2 i . e \tan^{4} x + \cot^{4} x = 194$

Page No 5.44:

Question 27:

If $\frac{3 π}{4} < x < π,$ then $\sqrt{{cosec}^{2} x + 2 \cot x}$ is equal to ___________.

Answer:

Given $\frac{3 π}{4} < x < π$
$\sqrt{{cosec}^{2} x + 2 \cot x} = \sqrt{2 \cot x + \cot^{2} x + 1} = \sqrt{{(\cot x + 1)}^{2}} = |\cot x + 1| Since \frac{3 π}{4} < x < π i . e \sqrt{{cosec}^{2} x + 2 \cot x} = - \cot x - 1$

Page No 5.44:

Question 1:

Write the maximum and minimum values of cos (cos x).

Answer:

$We know : - 1 \leq \cos x \leq 1 Also, \cos (- θ) = \cos θ When the angle increases from 0 to \frac{π}{2}, the value of \cos θ decreases . ∴ Maximum value of \cos [\cos (x)] = \cos (0) = 1' And, minimum value of \cos [\cos (x)] = c os (1)$

Page No 5.44:

Question 2:

Write the maximum and minimum values of sin (sin x).

Answer:

$We know : - 1 \leq \sin x \leq 1 Also, \sin (- θ) = - \sin θ When the angle increases from 0 to \frac{π}{2}, the value of \sin θ also increases . ∴ Maximum value of \sin [\sin (x)] = \sin (1) And, minimum value of \sin [\sin (x)] = \sin (- 1) = - \sin 1$

Page No 5.44:

Question 3:

Write the maximum value of sin (cos x).

Answer:

$We know : - 1 \leq \cos x \leq 1 Also, \sin (- θ) = - \sin θ W hen the angle increases from 0 to \frac{π}{2}, the value of \sin θ also increases . ∴ Maximum value of \sin [\cos x] = \sin (1)$

Page No 5.44:

Question 4:

If sin x = cos²x, then write the value of cos²x (1 + cos²x).

Answer:

$We have : \sin x = \cos^{2} x (1) ∴ \cos^{2} x (1 + \cos^{2} x) = \sin x (1 + \sin x) [Using (1)] = \sin x + \sin^{2} x = \sin x + 1 - \cos^{2} x = \sin x + 1 - \sin x [Using (1)] = 1$

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Question 5:

If sin x + cosec x = 2, then write the value of sinⁿ x + cosecⁿ x.

Answer:

$We have : \sin x + cosec x = 2 \Rightarrow \sin x + \frac{1}{\sin x} = 2 \Rightarrow \frac{\sin^{2} x + 1}{\sin x} = 2$ ^{$\Rightarrow \sin^{2} x + 1 = 2 \sin x \Rightarrow \sin^{2} x + 1 - 2 \sin x = 0 \Rightarrow {(\sin x - 1)}^{2} = 0 \Rightarrow \sin x - 1 = 0 \Rightarrow \sin x = 1 And, cosec x = \frac{1}{\sin x} = 1 ∴ \sin^{n} x + {cosec}^{n} x = 1^{n} + 1^{n} = 1 + 1 = 2$}

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Question 6:

If sin x + sin²x = 1, then write the value of cos¹²x + 3 cos¹⁰x + 3 cos⁸x + cos⁶x.

Answer:

$We have : \sin x + \sin^{2} x = 1 (1) \Rightarrow \sin x = 1 - \sin^{2} x \Rightarrow \sin x = c o s^{2} x (2) Now, taking cube of (1) : \sin x + \sin^{2} x = 1 \Rightarrow {(\sin x + \sin^{2} x)}^{3} = {(1)}^{3} \Rightarrow {(\sin x)}^{3} + {(\sin^{2} x)}^{3} + 3 {(\sin x)}^{2} (\sin^{2} x) + 3 (\sin x) {(\sin^{2} x)}^{2} = 1 \Rightarrow {(\sin x)}^{3} + {(\sin x)}^{6} + 3 {(\sin x)}^{4} + 3 {(\sin x)}^{5} = 1 \Rightarrow {(\sin x)}^{6} + 3 {(\sin x)}^{5} + 3 {(\sin x)}^{4} + {(\sin x)}^{3} = 1 \Rightarrow {(\cos^{2} x)}^{6} + 3 {(\cos^{2} x)}^{5} + 3 {(\cos^{2} x)}^{4} + {(\cos^{2} x)}^{3} = 1 \Rightarrow \cos^{12} x + 3 \cos^{10} x + 3 \cos^{8} x + \cos^{6} x = 1$

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Question 7:

If sin x + sin²x = 1, then write the value of cos⁸x + 2 cos⁶x + cos⁴x.

Answer:

$We have : \sin x + \sin^{2} x = 1 (1) \Rightarrow \sin x = 1 - \sin^{2} x \Rightarrow \sin x = c o s^{2} x (2) Now, taking square of (1) : \Rightarrow {(\sin x + \sin^{2} x)}^{2} = {(1)}^{2} \Rightarrow {(\sin x)}^{2} + {(\sin^{2} x)}^{2} + 2 (\sin x) (\sin^{2} x) = 1 \Rightarrow {(\sin x)}^{2} + {(\sin x)}^{4} + 2 {(\sin x)}^{3} = 1 \Rightarrow {(\sin x)}^{2} + 2 {(\sin x)}^{3} + {(\sin x)}^{4} = 1 \Rightarrow {({cos}^{2} x)}^{2} + 2 {({cos}^{2} x)}^{3} + {({cos}^{2} x)}^{4} = 1 \Rightarrow {cos}^{4} x + 2 {cos}^{6} x + {cos}^{8} x = 1 ∴ {cos}^{8} x + 2 {cos}^{6} x + {cos}^{4} x = 1$

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Question 8:

If sin θ₁ + sin θ₂ + sin θ₃ = 3, then write the value of cos θ₁+ cos θ₂ + cos θ₃.

Answer:

Sine function can take the maximum value of 1.
If, $\sin θ_{1} + \sin θ_{2} + \sin θ_{3} = 3$ , then we have:

sin $θ_{1}$ = 1

⇒ $θ_{1}$ = $\frac{π}{2}$
Similarly, $θ_{2} = θ_{3} = \frac{π}{2}$

$\Rightarrow \cos θ_{1} = \cos θ_{2} = \cos θ_{3} = 0 \Rightarrow \cos θ_{1} + \cos θ_{2} + \cos θ_{3} = 0$

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Question 9:

Write the value of sin 10° + sin 20° + sin 30° + ... + sin 360°.

Answer:

$\sin 10 ° + \sin 20 ° + . . . + \sin 170 ° + \sin 180 ° + \sin (360 ° - 170 °) + \sin (360 ° - 160 °) + . . . + \sin (360 ° - 20 °) + \sin (360 ° - 10 °) + \sin 360 ° = \sin 10 ° + \sin 20 ° + . . . + \sin 180 ° - \sin 170 ° - \sin 160 ° - . . . - \sin 20 ° - \sin 10 ° + \sin 360 ° (∵ \sin (360 ° - x) = - \sin x) = \sin 180 ° + \sin 360 ° = 0 + 0 = 0$

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Question 10:

A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm. Write the measure of the angle subtended by it at the centre of the loop.

Answer:

Circumference of the circle of radius 15 cm:
$2 πr = 2 \times 3.14 \times 15 cm = 94.2 cm$

Now, 94.2 cm will be the length of arc $(l)$ for the circle with radius 120 cm.
We know:
$l = r θ Here, θ is measured in radians . ∴ 94.2 = 120 \times θ \Rightarrow θ = 0.785 radians$

45 $°$ = $\frac{π}{4} = \frac{22}{7 \times 4} = 0.785$ radians
Therefore, the angle subtended by it at the centre of the loop is 45 $°$ .

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Question 11:

Write the value of 2 (sin⁶x + cos⁶ x) −3 (sin⁴ x + cos⁴x) + 1.

Answer:

$2 (\sin^{6} x + \cos^{6} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 = 2 (\sin^{2} x + \cos^{2} x) (\sin^{4} x + \cos^{4} x - \sin^{2} x . c {os}^{2} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 = 2.1 (\sin^{4} x + \cos^{4} x - \sin^{2} x . c {os}^{2} x) - 3 (\sin^{4} x + \cos^{4} x) + 1 = 2 (\sin^{4} x + \cos^{4} x) - 2 \sin^{2} x . c {os}^{2} x - 3 (\sin^{4} x + \cos^{4} x) + 1 = - (\sin^{4} x + \cos^{4} x) - 2 \sin^{2} x . c {os}^{2} x + 1 = - \{\sin^{4} x + \cos^{4} x + 2 \sin^{2} x . c {os}^{2} x\} + 1 = - {(\sin^{2} x + \cos^{2} x)}^{2} + 1 = - 1 + 1 = 0$

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Question 12:

Write the value of cos 1° + cos 2° + cos 3° + ... + cos 180°.

Answer:

$\cos 1 ° + \cos 2 ° + \cos 3 ° + . . . + \cos 180 ° = \cos 1 ° + \cos 2 ° + \cos 3 ° + . . . + \cos 88 ° + \cos 89 ° + \cos 90 ° + \cos (180 - 89) ° + \cos (180 - 88) ° + . . . + \cos (180 - 1) ° + \cos 180 ° [\cos (180 ° - θ) = - \cos θ] = \cos 1 ° + \cos 2 ° + \cos 3 ° + . . . + \cos 88 ° + \cos 89 ° + \cos 90 ° - \cos 89 ° - \cos 88 ° - . . . - \cos 1 ° + \cos 180 ° = \cos 90 ° + \cos 180 ° = 0 - 1 = - 1$

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Question 13:

If cot (α + β) = 0, then write the value of sin (α + 2β).

Answer:

$\cot (α + β) = 0 \Rightarrow α + β = \frac{π}{2} (1) β = \frac{π}{2} - α (2) α = \frac{π}{2} - β (3) Now, \sin (α + 2 β) = \sin (α + β + β) = \sin (\frac{π}{2} + \frac{π}{2} - α) = \sin (π - α) = \sin α Now, \sin (α + 2 β) = \sin (α + 2 β) = \sin (\frac{π}{2} - β + 2 β) = \sin (\frac{π}{2} + β) = \cos β$

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Question 14:

If tan A + cot A = 4, then write the value of tan⁴A + cot⁴A.

Answer:

$\tan A + \cot A = 4 Squaring both the sides : \tan^{2} A + c o t^{2} A + 2 = 16 \Rightarrow \tan^{2} A + c o t^{2} A = 14 Squaring both the sides again : \tan^{4} A + c o t^{4} A + 2 = 196 \Rightarrow \tan^{4} A + c o t^{4} A = 194$

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Question 15:

Write the least value of cos²x + sec²x.

Answer:

We know:
cos x can take the minimum value of $- 1$ .

cos²x + sec²x

$= \frac{\cos^{4} x + 1}{\cos^{2} x} = \frac{{(- 1)}^{4} + 1}{{(- 1)}^{2}} = 2$

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Question 16:

If x = sin¹⁴x + cos²⁰ x, then write the smallest interval in which the value of x lie.

Answer:

If x = 0 $°$ , 90 $°$ , 180 $°$ , 270 $°$ , 360 $°$ , then $\sin^{14} x + \cos^{20} x will always be 1 .$

The smallest interval in which the value of x lie is $(0, 1]$ .

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Question 17:

If 3 sin x + 5 cos x = 5, then write the value of 5 sin x − 3 cos x.

Answer:

$3 \sin x + 5 \cos x = 5 (Given) Squaring both the sides : 9 \sin^{2} x + 25 \cos^{2} x + 30 \sin x \cos x = 25 30 \sin x \cos x = 25 - 9 \sin^{2} x - 25 \cos^{2} x (1) We have to find the value of 5 \sin θ - 3 \cos θ . {(5 \sin x - 3 \cos x)}^{2} = 25 \sin^{2} x + 9 \cos^{2} x - 30 \sin x \cos x {(5 \sin x - 3 \cos x)}^{2} = 25 \sin^{2} x + 9 \cos^{2} x - (25 - 9 \sin^{2} x - 25 \cos^{2} x) [From (1)] {(5 \sin x - 3 \cos x)}^{2} = 34 \sin^{2} x + 34 \cos^{2} x - 25 {(5 \sin x - 3 \cos x)}^{2} = 34 - 25 (∵ \sin^{2} x + \cos^{2} x = 1) {(5 \sin x - 3 \cos x)}^{2} = 9 5 \sin x - 3 \cos x = \pm 3$

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