Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 11 T Ratios Of Some Particular Angles are provided here with simple step-by-step explanations. These solutions for T Ratios Of Some Particular Angles are extremely popular among Class 10 students for Maths T Ratios Of Some Particular Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 554:

$3\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{9}{3}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(9{x}^{2}-\frac{9}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{\left(3x\right)}^{2}-{\left(\frac{3}{x}\right)}^{2}\right]$
$=\frac{1}{3}\left[{\left(\mathrm{cosec}\theta \right)}^{2}-{\left(\mathrm{cot}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left({\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

Page No 554:

From the given right-angled triangle, we have:

Page No 554:

From the given right-angled triangle, we have:

Page No 554:

From right-angled ∆ABC, we have:

Page No 572:

On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

Ans

Page No 572:

On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
$=\left(\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

Page No 572:

On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o  sin 30o

As we know that,

Page No 572:

On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

As we know that,

Page No 572:

On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − $\frac{3}{4}$ sec2 45o + $\frac{1}{4}$ cosec2 30o

As we know that,

(i)

Hence, LHS = RHS

(ii)

Hence, LHS = RHS

1sin60°cos60°

Page No 572:

(i) sin 60o cos 30o − cos 60o sin 30o

∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o

∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
$=2×\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}=1$
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

Page No 572:

A = 45o
⇒ 2A = 2 $×$ 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =
Now, 1 − 2 sin2 A
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

Page No 572:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

​(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

=2tanA1+tan2A

Page No 572:

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o
∴​ cos (A + B) = cos A cos B − sin A sin B

Page No 573:

(i) sin (A − B) = sin 30o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o

∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴​ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = $\frac{1}{\sqrt{3}}$

∴​ tan (AB) =

Given:

13

Page No 573:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ tan 60o = $\sqrt{3}$

Page No 573:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ cos 30o = $\frac{\sqrt{3}}{2}$

Page No 573:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ sin 30o = $\frac{1}{2}$

Page No 573:

Disclaimer: $\mathrm{cos}15°$ can also be calculated by taking .

As we know that,

As we know that,

As we know that,

Page No 573:

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A B) = tan 30o       [∵ tan 30o = $\frac{1}{\sqrt{3}}$]
A − B = 30o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = $\sqrt{3}$]
A + B = 60o                           ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

As we know that,

Page No 574:

(i) As we know that,

(ii) As we know that,

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Page No 576:

Given: tan= 3cotx

As we know that,

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As we know that,