Rd Sharma XII Vol 1 2021 for Class 12 Science Maths Chapter 3 - Inverse Trigonometric Functions

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Rd Sharma XII Vol 1 2021 Solutions for Class 12 Science Maths Chapter 3 Inverse Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Inverse Trigonometric Functions are extremely popular among Class 12 Science students for Maths Inverse Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2021 Book of Class 12 Science Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2021 Solutions. All Rd Sharma XII Vol 1 2021 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 3.10:

Question 1:

Find the domain of definition of $f (x) = \cos^{- 1} (x^{2} - 4)$ .

Answer:

For $\cos^{- 1} (x^{2} - 4)$ to be defined
$- 1 \leq x^{2} - 4 \leq 1 \Rightarrow 3 \leq x^{2} \leq 5 \Rightarrow x \in [- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$
Hence, the domain of $f (x) is [- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$ .

Page No 3.10:

Question 2:

Find the domain of $f (x) = 2 \cos^{- 1} 2 x + \sin^{- 1} x$ .

Answer:

For $2 \cos^{- 1} 2 x$ to be defined.
$- 1 \leq 2 x \leq 1 \Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2} . . . . . (i)$
For $\sin^{- 1} x$ to be defined.
$- 1 \leq x \leq 1 . . . . . (ii)$
Domain of $f (x) = [- \frac{1}{2}, \frac{1}{2}] \cap [- 1, 1]$
$= [- \frac{1}{2}, \frac{1}{2}]$ .

Page No 3.10:

Question 3:

Find the domain of $f (x) = \cos^{- 1} x + \cos x$ .

Answer:

For $\cos^{- 1} x$ to be defined.
$- 1 \leq x \leq 1$
Now, cosx is defined for all real values.
So, domain of cosx is R.
Domain of $f (x) is R \cap [- 1, 1] = [- 1, 1]$ .

Page No 3.10:

Question 4:

Find the principal values of each of the following:

(i) $\cos^{- 1} (- \frac{\sqrt{3}}{2})$
(ii) $\cos^{- 1} (- \frac{1}{\sqrt{2}})$
(iii) $\cos^{- 1} (\sin \frac{4 π}{3})$

(iv) $\cos^{- 1} (\tan \frac{3 π}{4})$

Answer:

(i) Let $\cos^{- 1} (- \frac{\sqrt{3}}{2}) = y$
Then,
$\cos y = - \frac{\sqrt{3}}{2}$
We know that the range of the principal value branch is $[0, π]$ .
Thus,
$\cos y = - \frac{\sqrt{3}}{2} = \cos (\frac{5 π}{6}) \Rightarrow y = \frac{5 π}{6} \in [0, π]$
Hence, the principal value of $\cos^{- 1} (- \frac{\sqrt{3}}{2}) is \frac{5 π}{6}$ .

(ii) Let $\cos^{- 1} (- \frac{1}{\sqrt{2}}) = y$
Then,
$\cos y = - \frac{1}{\sqrt{2}}$
We know that the range of the principal value branch is $[0, π]$ .
Thus,
$\cos y = - \frac{1}{\sqrt{2}} = \cos (\frac{3 π}{4}) \Rightarrow y = \frac{3 π}{4} \in [0, π]$
Hence, the principal value of $\cos^{- 1} (- \frac{1}{\sqrt{2}}) is \frac{3 π}{4}$ .

(iii) Let $\cos^{- 1} (\sin \frac{4 π}{3}) = y$
Then,
$\cos y = \sin \frac{4 π}{3}$
We know that the range of the principal value branch is $[0, π]$ .
Thus,
$\cos y = \sin \frac{4 π}{3} = - \frac{\sqrt{3}}{2} = \cos (\frac{5 π}{6}) \Rightarrow y = \frac{5 π}{6} \in [0, π]$
Hence, the principal value of $\cos^{- 1} (\sin \frac{4 π}{3}) is \frac{5 π}{6}$ .

(iv) Let $\cos^{- 1} (\tan \frac{3 π}{4}) = y$
Then,
$\cos y = \tan \frac{3 π}{4}$
We know that the range of the principal value branch is $[0, π]$ .
Thus,
$\cos y = \tan \frac{3 π}{4} = - 1 = \cos (π) \Rightarrow y = π \in [0, π]$
Hence, the principal value of $\cos^{- 1} (\tan \frac{3 π}{4}) is π$ .

Page No 3.10:

Question 5:

For the principal values, evaluate each of the following:

(i) $\cos^{- 1} \frac{1}{2} + 2 \sin^{- 1} (\frac{1}{2})$
(ii)
(iii) $\sin^{- 1} (- \frac{1}{2}) + 2 \cos^{- 1} (- \frac{\sqrt{3}}{2})$
(iv) $\sin^{- 1} (- \frac{\sqrt{3}}{2}) + \cos^{- 1} (\frac{\sqrt{3}}{2})$

Answer:

$(i) \cos^{- 1} (\cos x) = x \sin^{- 1} (\sin x) = x$

$\cos^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2}) = \cos^{- 1} (\cos \frac{π}{3}) + 2 \sin^{- 1} (\sin \frac{π}{6}) = \frac{π}{3} + 2 (\frac{π}{6}) = \frac{2 π}{3}$

(ii)

$(iii) \sin^{- 1} (- \frac{1}{2}) + 2 \cos^{- 1} (- \frac{\sqrt{3}}{2}) = \sin^{- 1} \{\sin (- \frac{π}{6})\} + 2 \cos^{- 1} (\cos \frac{5 π}{6}) [∵ Range of sine is [- \frac{π}{2}, \frac{π}{2}]; - \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}] and range of cosine is [0, π]; \frac{5 π}{6} \in [0, π]] = - \frac{π}{6} + 2 (\frac{5 π}{6}) = - \frac{π}{6} + \frac{5 π}{3} = \frac{9 π}{6} = \frac{3 π}{2}$
$∴ \sin^{- 1} (- \frac{1}{2}) + 2 \cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{3 π}{2}$

$(iv) \sin^{- 1} (- \frac{\sqrt{3}}{2}) + \cos^{- 1} (\frac{\sqrt{3}}{2}) = \sin^{- 1} \{\sin (- \frac{π}{3})\} + \cos^{- 1} (\cos \frac{π}{6}) = - \frac{π}{3} + \frac{π}{6} [∵ Range of sine is [- \frac{π}{2}, \frac{π}{2}]; - \frac{π}{3} \in [- \frac{π}{2}, \frac{π}{2}] and range of cosine is [0, π]; \frac{π}{6} \in [0, π]] = - \frac{π}{6}$
$∴ \sin^{- 1} (- \frac{\sqrt{3}}{2}) + \cos^{- 1} (\frac{\sqrt{3}}{2}) = - \frac{π}{6}$

Page No 3.115:

Question 1:

Evaluate the following:

(i) $\tan \{2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}\}$
(ii) $\tan \frac{1}{2} (\cos^{- 1} \frac{\sqrt{5}}{3})$
(iii) $\sin (\frac{1}{2} \cos^{- 1} \frac{4}{5})$
(iv) $\sin (2 \tan^{- 1} \frac{2}{3}) + \cos (\tan^{- 1} \sqrt{3})$

Answer:

(i)

$\tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) = \tan (2 \tan^{- 1} \frac{1}{5} - \tan^{- 1} 1) = \tan [\tan^{- 1} \{\frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}}\} - \tan^{- 1} 1] [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan [\tan^{- 1} \{\frac{\frac{2}{5}}{\frac{24}{25}}\} - \tan^{- 1} 1] = \tan [\tan^{- 1} \frac{5}{12} + \tan^{- 1} 1] = \tan [\tan^{- 1} (\frac{\frac{5}{12} - 1}{1 + \frac{5}{12}})] [∵ \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})] = \tan [\tan^{- 1} (\frac{\frac{- 7}{12}}{\frac{17}{12}})] = \tan [\tan^{- 1} \frac{- 7}{17}] = \frac{- 7}{17}$

(ii)
$Let, \cos^{- 1} \frac{\sqrt{5}}{3} = θ \Rightarrow \cos θ = \frac{\sqrt{5}}{3} \Rightarrow 2 \cos^{2} \frac{θ}{2} - 1 = \frac{\sqrt{5}}{3} \Rightarrow \cos^{2} \frac{θ}{2} = \frac{3 + \sqrt{5}}{6} \Rightarrow \cos \frac{θ}{2} = \sqrt{\frac{3 + \sqrt{5}}{6}} \Rightarrow \frac{θ}{2} = \cos^{- 1} (\sqrt{\frac{3 + \sqrt{5}}{6}}) = \tan^{- 1} (\frac{\sqrt{1 - {(\sqrt{\frac{3 + \sqrt{5}}{6}})}^{2}}}{\sqrt{\frac{3 + \sqrt{5}}{6}}}) = \tan^{- 1} (\frac{\sqrt{1 - \frac{3 + \sqrt{5}}{6}}}{\sqrt{\frac{3 + \sqrt{5}}{6}}}) = \tan^{- 1} (\frac{\sqrt{\frac{3 - \sqrt{5}}{6}}}{\sqrt{\frac{3 + \sqrt{5}}{6}}}) = \tan^{- 1} (\sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}) = \tan^{- 1} (\sqrt{\frac{(3 - \sqrt{5}) (3 - \sqrt{5})}{(3 + \sqrt{5}) (3 - \sqrt{5})}}) = \tan^{- 1} (\sqrt{\frac{{(3 - \sqrt{5})}^{2}}{9 - 5}}) = \tan^{- 1} (\frac{3 - \sqrt{5}}{2}) i . e ., \frac{1}{2} (\cos^{- 1} \frac{\sqrt{5}}{3}) = \tan^{- 1} (\frac{3 - \sqrt{5}}{2}) \Rightarrow \tan \frac{1}{2} (\cos^{- 1} \frac{\sqrt{5}}{3}) = \tan [\tan^{- 1} (\frac{3 - \sqrt{5}}{2})] ∴ \tan \frac{1}{2} (\cos^{- 1} \frac{\sqrt{5}}{3}) = \frac{3 - \sqrt{5}}{2}$

(iii)
$\sin (\frac{1}{2} \cos^{- 1} \frac{4}{5}) = \sin \{\frac{1}{2} \times 2 \sin^{- 1} \pm \sqrt{\frac{1 - \frac{4}{5}}{2}}\} [∵ \cos^{- 1} x = 2 \sin^{- 1} \pm \sqrt{\frac{1 - x}{2}}] = \sin (\sin^{- 1} \pm \frac{1}{\sqrt{10}}) = \pm \frac{1}{\sqrt{10}}$

(iv)
$\sin (2 \tan^{- 1} \frac{2}{3}) + \cos (\tan^{- 1} \sqrt{3}) = \sin (\sin^{- 1} \frac{2 \times \frac{2}{3}}{1 + \frac{4}{9}}) + \cos (\cos^{- 1} \frac{1}{\sqrt{1 + {(\sqrt{3})}^{2}}}) = \sin (\sin^{- 1} \frac{12}{13}) + \cos (\cos^{- 1} \frac{1}{2}) = \frac{12}{13} + \frac{1}{2} = \frac{37}{26}$

Page No 3.115:

Question 2:

(i) $2 \sin^{- 1} \frac{3}{5} = \tan^{- 1} \frac{24}{7}$
(ii) $\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \frac{1}{2} \cos^{- 1} \frac{3}{5} = \frac{1}{2} \sin^{- 1} (\frac{4}{5})$
(iii) $\tan^{- 1} \frac{2}{3} = \frac{1}{2} \tan^{- 1} \frac{12}{5}$

(iv) $\tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{4}$

(v) $\sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{2}$
(vi) $2 \sin^{- 1} \frac{3}{5} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}$
(vii) $2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \frac{4}{7}$
(viii) $2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}$
(ix) $2 \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) = \tan^{- 1} (\frac{31}{17})$
(x) $4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239} = \frac{π}{4}$

Answer:

$(i) LHS = 2 \sin^{- 1} \frac{3}{5} = 2 \tan^{- 1} \frac{\frac{3}{4}}{\sqrt{1 - \frac{9}{25}}} [∵ \sin^{- 1} x = \tan^{- 1} \frac{x}{\sqrt{1 - x^{2}}}] = 2 \tan^{- 1} \frac{\frac{3}{5}}{\frac{4}{5}} = 2 \tan^{- 1} \frac{3}{4} = \tan^{- 1} \{\frac{2 \times \frac{3}{4}}{1 - {(\frac{3}{4})}^{2}}\} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \{\frac{\frac{3}{2}}{\frac{7}{16}}\} = \tan^{- 1} = RHS$

$(ii) LHS = \tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \tan^{- 1} (\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \times \frac{2}{9}}) [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})] = \tan^{- 1} (\frac{\frac{17}{36}}{\frac{34}{36}}) = \tan^{- 1} \frac{1}{2} = \frac{1}{2} \cos^{- 1} (\frac{1 - \frac{1}{4}}{1 + \frac{1}{4}}) [∵ \tan^{- 1} x = \frac{1}{2} \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] = \frac{1}{2} \cos^{- 1} (\frac{\frac{3}{4}}{\frac{5}{4}}) = \frac{1}{2} \cos^{- 1} (\frac{3}{5}) Now, \tan^{- 1} \frac{1}{2} = \frac{1}{2} \sin^{- 1} (\frac{\frac{2}{2}}{1 + \frac{1}{4}}) [∵ \tan^{- 1} x = \frac{1}{2} \sin^{- 1} (\frac{2 x}{1 + x^{2}})] = \frac{1}{2} \sin^{- 1} (\frac{1}{\frac{5}{4}}) = \frac{1}{2} \sin^{- 1} (\frac{4}{5})$

$(iii) LHS = \tan^{- 1} \frac{2}{3} = \frac{1}{2} \tan^{- 1} \{\frac{2 \times \frac{2}{3}}{1 - {(\frac{2}{3})}^{2}}\} [∵ \tan^{- 1} x = \frac{1}{2} \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \frac{1}{2} \tan^{- 1} \{\frac{\frac{4}{3}}{\frac{5}{9}}\} = \frac{1}{2} \tan^{- 1} \frac{12}{5} = RHS$

$(iv) LHS = \tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3} = \tan^{- 1} \frac{1}{7} + \tan^{- 1} \{\frac{2 \times \frac{1}{3}}{1 - {(\frac{1}{3})}^{2}}\} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \frac{1}{7} + \tan^{- 1} \{\frac{\frac{2}{3}}{\frac{8}{9}}\} = \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{3}{4} = \tan^{- 1} (\frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \times \frac{3}{4}}) [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})] = \tan^{- 1} (\frac{\frac{25}{28}}{\frac{25}{28}}) = \tan^{- 1} 1 = \frac{π}{4} = RHS$

$(v) LHS = \sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} = \sin^{- 1} \frac{4}{5} + \tan^{- 1} \{\frac{2 \times \frac{1}{3}}{1 - {(\frac{1}{3})}^{2}}\} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \sin^{- 1} \frac{4}{5} + \tan^{- 1} \{\frac{\frac{2}{3}}{\frac{8}{9}}\} = \sin^{- 1} \frac{4}{5} + \tan^{- 1} \frac{3}{4} = \sin^{- 1} \frac{4}{5} + \cos^{- 1} \frac{1}{\sqrt{1 + \frac{9}{16}}} [∵ \tan^{- 1} x = \cos^{- 1} \frac{1}{\sqrt{1 + x^{2}}}] = \sin^{- 1} \frac{4}{5} + \cos^{- 1} \frac{1}{\frac{5}{4}} = \sin^{- 1} \frac{4}{5} + \cos^{- 1} \frac{4}{5} = \frac{π}{2} = RHS$

$(vi) LHS = 2 \sin^{- 1} \frac{3}{5} - \tan^{- 1} \frac{17}{31} = 2 \tan^{- 1} \frac{\frac{3}{4}}{\sqrt{1 - \frac{9}{25}}} - \tan^{- 1} \frac{17}{31} [∵ \sin^{- 1} x = \tan^{- 1} \frac{x}{\sqrt{1 - x^{2}}}] = 2 \tan^{- 1} \frac{\frac{3}{5}}{\frac{4}{5}} - \tan^{- 1} \frac{17}{31} = 2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \tan^{- 1} \{\frac{2 \times \frac{3}{4}}{1 - {(\frac{3}{4})}^{2}}\} - \tan^{- 1} \frac{17}{31} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \{\frac{\frac{3}{2}}{\frac{7}{16}}\} - \tan^{- 1} \frac{17}{31} = \tan^{- 1} \frac{24}{7} - \tan^{- 1} \frac{17}{31} = \tan^{- 1} (\frac{\frac{24}{7} - \frac{17}{31}}{1 + \frac{24}{7} \times \frac{17}{31}}) [∵ \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})] = \tan^{- 1} (\frac{\frac{625}{217}}{\frac{625}{217}}) = \tan^{- 1} 1 = \frac{π}{4} = RHS$

$(vii) LHS = 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \{\frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}}\} + \tan^{- 1} \frac{1}{8} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \{\frac{\frac{2}{5}}{\frac{24}{25}}\} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \frac{5}{12} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} (\frac{\frac{5}{12} + \frac{1}{8}}{1 - \frac{5}{12} \times \frac{1}{8}}) [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})] = \tan^{- 1} (\frac{\frac{13}{24}}{\frac{91}{96}}) = \tan^{- 1} \frac{4}{7} = RHS$

$(viii) LHS = 2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \tan^{- 1} \{\frac{2 \times \frac{3}{4}}{1 - {(\frac{3}{4})}^{2}}\} - \tan^{- 1} \frac{17}{31} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \{\frac{\frac{3}{2}}{\frac{7}{16}}\} - \tan^{- 1} \frac{17}{31} = \tan^{- 1} \frac{24}{7} - \tan^{- 1} \frac{17}{31} = \tan^{- 1} (\frac{\frac{24}{7} - \frac{17}{31}}{1 + \frac{24}{7} \times \frac{17}{31}}) [∵ \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})] = \tan^{- 1} (\frac{\frac{625}{217}}{\frac{625}{217}}) = \tan^{- 1} 1 = \frac{π}{4} = RHS$

$(ix) LHS = 2 \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{7} = \tan^{- 1} \{\frac{2 \times \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}}\} + \tan^{- 1} \frac{1}{7} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \{\frac{1}{\frac{3}{4}}\} + \tan^{- 1} \frac{1}{7} = \tan^{- 1} \frac{4}{3} + \tan^{- 1} \frac{1}{7} = \tan^{- 1} (\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}}) [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})] = \tan^{- 1} (\frac{\frac{31}{21}}{\frac{17}{21}}) = \tan^{- 1} \frac{31}{17} = RHS$

$(x) LHS = 4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239} = 2 \tan^{- 1} \{\frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}}\} - \tan^{- 1} \frac{1}{239} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = 2 \tan^{- 1} \{\frac{\frac{2}{5}}{\frac{24}{25}}\} - \tan^{- 1} \frac{1}{239} = 2 \tan^{- 1} \frac{5}{12} - \tan^{- 1} \frac{1}{239} = \tan^{- 1} \{\frac{2 \times \frac{5}{12}}{1 - {(\frac{5}{12})}^{2}}\} - \tan^{- 1} \frac{1}{239} [∵ 2 \tan^{- 1} x = \tan^{- 1} \{\frac{2 x}{1 - x^{2}}\}] = \tan^{- 1} \{\frac{\frac{5}{6}}{\frac{119}{144}}\} - \tan^{- 1} \frac{1}{239} = \tan^{- 1} \frac{120}{119} - \tan^{- 1} \frac{1}{239} = \tan^{- 1} (\frac{\frac{120}{119} - \frac{17}{239}}{1 + \frac{120}{119} \times \frac{1}{239}}) [∵ \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})] = \tan^{- 1} 1 = \frac{π}{4} = RHS$

Page No 3.115:

Question 3:

$If \sin^{- 1} \frac{2 a}{1 + a^{2}} - \cos^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1} \frac{2 x}{1 - x^{2}}, then prove that x = \frac{a - b}{1 + a b}$

Answer:

$Let : a = \tan m b = \tan n x = \tan y$

Now,
$\sin^{- 1} \frac{2 a}{1 + a^{2}} - \cos^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1} \frac{2 x}{1 - x^{2}} \Rightarrow \sin^{- 1} \frac{2 \tan m}{1 + \tan^{2} m} - \cos^{- 1} \frac{1 - \tan^{2} n}{1 + \tan^{2} n} = \tan^{- 1} \frac{2 \tan y}{1 - \tan^{2} y} \Rightarrow \sin^{- 1} (\sin 2 m) - \cos^{- 1} (\cos 2 n) = \tan^{- 1} (\tan 2 y) [∵ \sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x} and \cos 2 x = \frac{1 - \tan^{2} x}{1 + \tan^{2} x}] \Rightarrow 2 m - 2 n = 2 y \Rightarrow m - n = y \Rightarrow \tan^{- 1} a - \tan^{- 1} b = \tan^{- 1} x [∵ a = \tan m, b = \tan n and x = \tan y] \Rightarrow \tan^{- 1} \frac{a - b}{1 + a b} = \tan^{- 1} x [∵ \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} \frac{x - y}{1 + x y}] \Rightarrow \frac{a - b}{1 + a b} = x$

$∴ \frac{a - b}{1 + a b} = x$

Page No 3.115:

Question 4:

Prove that
(i) $\tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{2}$

(ii) $\sin \{\tan^{- 1} \frac{1 - x^{2}}{2 x} + \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}}\} = 1$
(iii) $\sin^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \cos^{- 1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

Answer:

(i)
$\tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{2} LHS = \tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \frac{π}{2} - \tan^{- 1} (\frac{1 - x^{2}}{2 x}) [∵ \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}] = \frac{π}{2} = RHS$

(ii)
$\sin \{\tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})\} = 1 LHS = \sin \{\tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})\} = \sin \{\sin^{- 1} (\frac{\frac{1 - x^{2}}{2 x}}{\sqrt{1 + \frac{1 - x^{2}}{2 x}}}) + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})\} [∵ \tan^{- 1} x = \sin^{- 1} \frac{x}{\sqrt{1 + x^{2}}}] = \sin \{\sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})\} = \sin \{\frac{π}{2}\} [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}] = 1 = RHS$
(iii)
To prove:
$\sin^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \cos^{- 1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$
Let us consider $\cos^{- 1} (x) = θ$
$\Rightarrow x = \cos θ$
Taking R.H.S.
$\sin^{- 1} (2 \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow \sin^{- 1} (\sin 2 θ) (∵ 2 \sin θ \cos θ = \sin 2 θ) \Rightarrow 2 θ \Rightarrow 2 \cos^{- 1} x = RHS (∵ θ = \cos^{- 1} x)$
Hence, proved.

Page No 3.115:

Question 5:

$If \sin^{- 1} \frac{2 a}{1 + a^{2}} + \sin^{- 1} \frac{2 b}{1 + b^{2}} = 2 \tan^{- 1} x, prove that x = \frac{a + b}{1 - a b} .$

Answer:

Let:
$a = \tan z b = \tan y$

Then,
$\sin^{- 1} \frac{2 a}{1 + a^{2}} + \sin^{- 1} \frac{2 b}{1 + b^{2}} = 2 \tan^{- 1} x \Rightarrow \sin^{- 1} \frac{2 \tan z}{1 + \tan^{2} z} + \sin^{- 1} \frac{2 \tan y}{1 + \tan^{2} y} = 2 \tan^{- 1} x \Rightarrow \sin^{- 1} (\sin 2 z) + \sin^{- 1} (\sin 2 y) = 2 \tan^{- 1} x [∵ \sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x}] \Rightarrow 2 z + 2 y = 2 \tan^{- 1} x \Rightarrow \tan^{- 1} a + \tan^{- 1} b = \tan^{- 1} x [∵ a = \tan z and b = \tan y] \Rightarrow \tan^{- 1} \frac{a + b}{1 - a b} = \tan^{- 1} x [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}] \Rightarrow x = \frac{a + b}{1 - a b}$

Page No 3.115:

Question 6:

Show that 2 tan⁻¹ x + sin⁻¹ $\frac{2 x}{1 + x^{2}}$ is constant for x ≥ 1, find that constant.

Answer:

We have
$2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) (1) For x > 1, = 2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}) + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) [∵ 2 \tan^{- 1} x = π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}), x > 1] = π (2) For x = 1, = 2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 \tan^{- 1} (1) + \sin^{- 1} (\frac{2 (1)}{1 + {(1)}^{2}}) = 2 \tan^{- 1} (1) + \sin^{- 1} (1) = 2 (\frac{π}{4}) + \frac{π}{2} = π$

Page No 3.116:

Question 7:

Find the values of each of the following:
(i) $\tan^{- 1} \{2 \cos (2 \sin^{- 1} \frac{1}{2})\}$
(ii) $\cos (\sec^{- 1} x + {cosec}^{- 1} x), |x| \geq 1$

Answer:

(i) Let $\sin^{- 1} \frac{1}{2} = y$
Then,
$\sin y = \frac{1}{2}$

$∴ \tan^{- 1} \{2 \cos (2 \sin^{- 1} \frac{1}{2})\} = \tan^{- 1} \{2 \cos 2 y\} = \tan^{- 1} (2 (1 - 2 \sin^{2} y)) [∵ \cos 2 x = 1 - 2 \sin^{2} x] = \tan^{- 1} \{2 (1 - 2 \times \frac{1}{4})\} [∵ \sin y = \frac{1}{2}] = \tan^{- 1} \{2 \times \frac{1}{2}\} = \tan^{- 1} 1 = \frac{π}{4}$

$∴ \tan^{- 1} \{2 \cos (2 \sin^{- 1} \frac{1}{2})\} = \frac{π}{4}$
(ii)
We have
$\cos (\sec^{- 1} x + {cosec}^{- 1} x) = \cos \frac{π}{2} [∵ \sec^{- 1} x + {cosec}^{- 1} x = \frac{π}{2}] = 0$

$∴ \cos (\sec^{- 1} x + {cosec}^{- 1} x) = 0, | x | \geq 1$

Page No 3.116:

Question 8:

Solve the following equations for x:

(i) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(ii) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(iii) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(iv) 2 $\tan^{- 1}$ ( $\sin$ $x$ ) $=$ $\tan$ ^{$-$ 1} (2 $\sin$ $x$ ), $x \neq \frac{π}{2}$ .

(v) $\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}$

(vi) $\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}$

Answer:

(i) We know
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$

$∴ \tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{\frac{1}{4} + \frac{1}{5}}{1 - \frac{1}{4} \times \frac{1}{5}}) + \tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{6}}{1 - \frac{1}{5} \times \frac{1}{6}}) + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{\frac{9}{20}}{\frac{19}{20}}) + \tan^{- 1} (\frac{\frac{11}{30}}{\frac{29}{30}}) + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{9}{19}) + \tan^{- 1} (\frac{11}{29}) + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{\frac{9}{19} + \frac{11}{29}}{1 - \frac{11}{29} \times \frac{9}{19}}) + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{235}{226}) + \tan^{- 1} \frac{1}{x} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{\frac{235}{226} + \frac{1}{x}}{1 - \frac{235}{226} \times \frac{1}{x}}) = \frac{π}{4} \Rightarrow \frac{235 x + 226}{226 x - 235} = \tan \frac{π}{4} \Rightarrow \frac{235 x + 226}{226 x - 235} = 1 \Rightarrow 235 x + 226 = 226 x - 235 \Rightarrow 9 x = - 461 \Rightarrow x = - \frac{461}{9}$

(ii)

$3 \sin^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3} \Rightarrow 6 \tan^{- 1} x - 8 \tan^{- 1} x + 4 \tan^{- 1} x = \frac{π}{3} [∵ 2 \tan^{- 1} x = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), 2 \tan^{- 1} x = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) and 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})] \Rightarrow 2 \tan^{- 1} x = \frac{π}{3} \Rightarrow \tan^{- 1} x = \frac{π}{6} \Rightarrow x = \tan \frac{π}{6} \Rightarrow x = \frac{1}{\sqrt{3}}$

(iii) We know
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$

$∴ \tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3} \Rightarrow \tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3} [∵ \cot^{- 1} x = \tan^{- 1} \frac{1}{x}] \Rightarrow \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3} \Rightarrow 2 \tan^{- 1} x = \frac{π}{3} [∵ 2 \tan^{- 1} x \tan^{- 1} (\frac{2 x}{1 - x^{2}})] \Rightarrow \tan^{- 1} x = \frac{π}{6} \Rightarrow x = \tan \frac{π}{6} \Rightarrow x = \frac{1}{\sqrt{3}}$

(iv) 2 $\tan^{- 1}$ ( $\sin$ $x$ ) $=$ $\tan$ ^{$-$ 1} (2 $\sin$ $x$ ), $x \neq \frac{π}{2}$

$\Rightarrow \tan^{- 1} (\frac{2 \sin x}{1 - \sin^{2} x}) = \tan^{- 1} (2 \sin x) [∵ 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})] \Rightarrow \frac{2 \sin x}{1 - \sin^{2} x} = 2 \sin x \Rightarrow 2 \sin x = 2 \sin x - 2 \sin^{3} x \Rightarrow 2 \sin^{3} x = 0 \Rightarrow \sin x = 0 \Rightarrow x = 0$

(v) $\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3} \Rightarrow \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + \frac{1}{2} \times 2 \tan^{- 1} x = \frac{2 π}{3} [∵ \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 \tan^{- 1} x] \Rightarrow 2 \tan^{- 1} x + \tan^{- 1} x = \frac{2 π}{3} [∵ \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 \tan^{- 1} x] \Rightarrow 3 \tan^{- 1} x = \frac{2 π}{3} \Rightarrow \tan^{- 1} x = \frac{2 π}{9} \Rightarrow x = \tan (\frac{2 π}{9})$

(vi)
$\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \tan^{- 1} 1 \Rightarrow \tan^{- 1} (\frac{x - 2}{x - 1}) = \tan^{- 1} 1 - \tan^{- 1} (\frac{x + 2}{x + 1}) \Rightarrow \tan^{- 1} (\frac{x - 2}{x - 1}) = \tan^{- 1} (\frac{1 - \frac{x + 2}{x + 1}}{1 + \frac{x + 2}{x + 1}}) \Rightarrow \tan^{- 1} (\frac{x - 2}{x - 1}) = \tan^{- 1} (\frac{x + 1 - x - 2}{x + 1 + x + 2}) \Rightarrow \tan^{- 1} (\frac{x - 2}{x - 1}) = \tan^{- 1} (\frac{- 1}{2 x + 3}) \Rightarrow \frac{x - 2}{x - 1} = \frac{- 1}{2 x + 3} \Rightarrow 2 x^{2} + 3 x - 4 x - 6 = - x + 1 \Rightarrow 2 x^{2} = 1 + 6 \Rightarrow x^{2} = 7 \Rightarrow x = \pm \sqrt{\frac{7}{2}}$

Page No 3.116:

Question 9:

$Prove that 2 \tan^{- 1} (\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}) = \cos^{- 1} (\frac{a \cos θ + b}{a + b \cos θ})$

Answer:

$LHS = 2 \tan^{- 1} (\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}) = \cos^{- 1} \{\frac{1 - {(\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2})}^{2}}{1 + {(\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2})}^{2}}\} [∵ 2 \tan^{- 1} (x) = \cos^{- 1} \{\frac{1 - x^{2}}{1 + x^{2}}\}] = \cos^{- 1} \{\frac{1 - \frac{a - b}{a + b} \tan^{2} \frac{θ}{2}}{1 + \frac{a - b}{a + b} \tan^{2} \frac{θ}{2}}\} = \cos^{- 1} \{\frac{a + b - (a - b) \tan^{2} \frac{θ}{2}}{a + b + (a - b) \tan^{2} \frac{θ}{2}}\} = \cos^{- 1} \{\frac{a + b - a \tan^{2} \frac{θ}{2} + b \tan^{2} \frac{θ}{2}}{a + b + a \tan^{2} \frac{θ}{2} - b \tan^{2} \frac{θ}{2}}\} = \cos^{- 1} \{\frac{a (1 - \tan^{2} \frac{θ}{2}) + b (1 + \tan^{2} \frac{θ}{2})}{a (1 + \tan^{2} \frac{θ}{2}) + b (1 - \tan^{2} \frac{θ}{2})}\} = \cos^{- 1} \{\frac{a (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + b (\frac{1 + \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}})}{a (\frac{1 + \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + b (\frac{1 - \tan^{2} \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}})}\} [Dividing N^{r} and D^{r} by 1 + \tan^{2} \frac{θ}{2}] = \cos^{- 1} \{\frac{a (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + b}{a + b (\frac{1 - \tan^{2} \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}})}\} = \cos^{- 1} \{\frac{a \cos θ + b}{a + b \cos θ}\} = RHS$

Page No 3.116:

Question 10:

Prove that:
$\tan^{- 1} \frac{2 a b}{a^{2} - b^{2}} + \tan^{- 1} \frac{2 x y}{x^{2} - y^{2}} = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}},$
where α = ax − by and β = ay + bx.

Answer:

We know
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y}), x y > 1$

$∴ \tan^{- 1} \frac{2 a b}{a^{2} - b^{2}} + \tan^{- 1} \frac{2 x y}{x^{2} - y^{2}} = \tan^{- 1} (\frac{\frac{2 a b}{a^{2} - b^{2}} + \frac{2 x y}{x^{2} - y^{2}}}{1 - \frac{2 a b}{a^{2} - b^{2}} \frac{2 x y}{x^{2} - y^{2}}}) = \tan^{- 1} (\frac{\frac{2 (a b x^{2} - a b y^{2} + x y a^{2} - x y b^{2})}{(a^{2} - b^{2}) (x^{2} - y^{2})}}{\frac{(a^{2} x^{2} - a^{2} y^{2} - x^{2} b^{2} + y^{2} b^{2} - 4 a b x y)}{(a^{2} - b^{2}) (x^{2} - y^{2})}}) = \tan^{- 1} (\frac{2 (a b x^{2} - a b y^{2} + x y a^{2} - x y b^{2})}{(a^{2} x^{2} - a^{2} y^{2} - x^{2} b^{2} + y^{2} b^{2} - 2 a b x y - 2 a b x y)}) = \tan^{- 1} (\frac{2 (a x - b y) (a y + b x)}{{(a x - b y)}^{2} - {(a y + b x)}^{2}}) = \tan^{- 1} (\frac{2 α β}{α^{2} - β^{2}}) [∵ α = a x - b y and β = a y + b x]$

Page No 3.116:

Question 11:

For any a, b, x, y > 0, prove that:
$\frac{2}{3} \tan^{- 1} (\frac{3 a b^{2} - a^{3}}{b^{3} - 3 a^{2} b}) + \frac{2}{3} \tan^{- 1} (\frac{3 x y^{2} - x^{3}}{y^{3} - 3 x^{2} y}) = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}}$
where α = − ax + by, β = bx + ay

Answer:

$Let a = b \tan m and x = y \tan n$
Then,

$\frac{2}{3} \tan^{- 1} (\frac{3 a b^{2} - a^{3}}{b^{3} - 3 a^{2} b}) + \frac{2}{3} \tan^{- 1} (\frac{3 x y^{2} - x^{3}}{y^{3} - 3 x^{2} y}) = \frac{2}{3} \tan^{- 1} (\frac{3 b^{3} \tan m - b^{3} \tan^{3} m}{b^{3} - 3 b^{3} \tan^{2} m}) + \frac{2}{3} \tan^{- 1} (\frac{3 y^{3} \tan n - y^{3} \tan^{3} n}{y^{3} - 3 y^{3} \tan^{2} n}) = \frac{2}{3} \tan^{- 1} (\frac{3 \tan m - \tan^{3} m}{1 - 3 \tan^{2} m}) + \frac{2}{3} \tan^{- 1} (\frac{3 \tan n - \tan^{3} n}{1 - \tan^{2} n}) = \frac{2}{3} \tan^{- 1} (\tan 3 m) + \frac{2}{3} \tan^{- 1} (\tan 3 n) [∵ \tan 3 x = \frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x}] = \frac{2}{3} (3 m) + \frac{2}{3} (3 n) = 2 m + 2 n = 2 (\tan^{- 1} \frac{a}{b} + \tan^{- 1} \frac{x}{y}) [∵ a = b \tan m, x = y \tan n] = 2 \tan^{- 1} (\frac{\frac{a}{b} + \frac{x}{y}}{1 - \frac{a}{b} \frac{x}{y}}) = 2 \tan^{- 1} (\frac{a y + b x}{b y - a x}) = \tan^{- 1} \{\frac{2 \frac{a y + b x}{b y - a x}}{1 - {(\frac{a y + b x}{b y - a x})}^{2}}\} = \tan^{- 1} \{\frac{2 (a y + b x) (b y - a x)}{{(b y - a x)}^{2} - {(a y + b x)}^{2}}\} = \tan^{- 1} \{\frac{2 α β}{α^{2} - β^{2}}\} [∵ β = a y + b x and α = b y - a x]$

Page No 3.116:

Question 1:

If $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}})$ = α, then x² =
(a) sin 2 α
(b) sin α
(c) cos 2 α
(d) cos α

Answer:

(a) sin 2α

$\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}) = α \Rightarrow \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}} = \tan α \Rightarrow \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}} \times \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}} = \tan α \Rightarrow \frac{{(\sqrt{1 + x^{2}})}^{2} + {(\sqrt{1 - x^{2}})}^{2} - 2 \sqrt{1 + x^{2}} \sqrt{1 - x^{2}}}{{(\sqrt{1 + x^{2}})}^{2} - {(\sqrt{1 - x^{2}})}^{2}} = \tan α \Rightarrow \frac{1 - \sqrt{1 - x^{4}}}{x^{2}} = \tan α \Rightarrow x^{2} \tan α = 1 - \sqrt{1 - x^{4}} \Rightarrow \sqrt{1 - x^{4}} = 1 - x^{2} \tan α \Rightarrow 1 - x^{4} = 1 + x^{4} \tan^{2} α - 2 x^{2} \tan α \Rightarrow x^{4} + x^{4} \tan^{2} α - 2 x^{2} \tan α = 0 \Rightarrow x^{4} \sec^{2} α - 2 x^{2} \tan α = 0 \Rightarrow x^{2} (x^{2} \sec^{2} α - 2 \tan α) = 0 \Rightarrow x^{2} \sec^{2} α - 2 \tan α = 0 [∵ x^{2} \neq 0] \Rightarrow x^{2} \sec^{2} α = 2 \tan α \Rightarrow x^{2} = \frac{2 \tan α}{\sec^{2} α} = 2 \sin α \cos α = \sin 2 α$

Page No 3.116:

Question 2:

The value of tan $\{\cos^{- 1} \frac{1}{5 \sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}}\}$ is
(a) $\frac{\sqrt{29}}{3}$

(b) $\frac{29}{3}$

(c) $\frac{\sqrt{3}}{29}$

(d) $\frac{3}{29}$

Answer:

(d) $\frac{3}{29}$

$Let, \cos^{- 1} \frac{1}{5 \sqrt{2}} = y and \sin^{- 1} \frac{4}{\sqrt{17}} = z$

$∴ \cos y = \frac{1}{5 \sqrt{2}} \Rightarrow \sin y = \frac{7}{5 \sqrt{2}} \Rightarrow \tan y = 7 \sin z = \frac{4}{\sqrt{17}} \Rightarrow \cos z = \frac{1}{\sqrt{17}} \Rightarrow \tan z = 4$

$∴ \tan (\cos^{- 1} \frac{1}{5 \sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}}) = \tan (y - z) = \frac{\tan y - \tan z}{1 + \tan y \tan z} = \frac{7 - 4}{1 + 7 \times 4} = \frac{3}{29}$

Page No 3.117:

Question 3:

2 tan⁻¹ {cosec (tan⁻¹ x) − tan (cot⁻¹ x)} is equal to
(a) cot⁻¹ x

(b) cot⁻¹ $\frac{1}{x}$

(c) tan⁻¹ x

(d) none of these

Answer:

(c) tan⁻¹ x

Let $\tan^{- 1} x = y$
So, $x = \tan y$

$∴ 2 \tan^{- 1} \{cosec (\tan^{- 1} x) - \tan (c o t^{- 1} x)\} = 2 \tan^{- 1} \{cosec (\tan^{- 1} x) - \tan (\tan^{- 1} \frac{1}{x})\} = 2 \tan^{- 1} \{cosec (\tan^{- 1} x) - \frac{1}{x}\} = 2 \tan^{- 1} \{cosec y - \frac{1}{\tan y}\} = 2 \tan^{- 1} \{\frac{1 - \cos y}{\sin y}\} = 2 \tan^{- 1} \{\frac{2 \sin^{2} \frac{y}{2}}{\sin y}\} = 2 \tan^{- 1} \{\frac{2 \sin^{2} \frac{y}{2}}{2 \sin \frac{y}{2} \cos \frac{y}{2}}\} = 2 \tan^{- 1} \{\tan \frac{y}{2}\} = y = \tan^{- 1} x$

Page No 3.117:

Question 4:

If $\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = α, then \frac{x^{2}}{a^{2}} - \frac{2 x y}{a b} \cos α + \frac{y^{2}}{b^{2}} =$
(a) sin² α
(b) cos² α
(c) tan² α
(d) cot² α

Answer:

(a) sin² α

We know that $\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} (x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}})$ .

$∴ \cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = α \Rightarrow \cos^{- 1} (\frac{x}{a} \frac{y}{b} - \sqrt{1 - \frac{x^{2}}{a^{2}}} \sqrt{1 - \frac{y^{2}}{b^{2}}}) = α \Rightarrow \frac{x y}{a b} - \sqrt{1 - \frac{x^{2}}{a^{2}}} \sqrt{1 - \frac{y^{2}}{b^{2}}} = \cos α \Rightarrow \sqrt{1 - \frac{x^{2}}{a^{2}}} \sqrt{1 - \frac{y^{2}}{b^{2}}} = \frac{x y}{a b} - \cos α \Rightarrow (1 - \frac{x^{2}}{a^{2}}) (1 - \frac{y^{2}}{b^{2}}) = \frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}} + \cos^{2} α - \frac{2 x y}{a b} \cos α [Squaring both the sides] \Rightarrow 1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}} = \frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}} + \cos^{2} α - \frac{2 x y}{a b} \cos α \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{2 x y}{a b} \cos α = 1 - \cos^{2} α = \sin^{2} α$

Page No 3.117:

Question 5:

The positive integral solution of the equation
$\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^{2}}} = \sin^{- 1} \frac{3}{\sqrt{10}} is$
(a) x = 1, y = 2
(b) x = 2, y = 1
(c) x = 3, y = 2
(d) x = −2, y = −1.

Answer:

(a) x = 1, y = 2

$We have, \tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^{2}}} = \sin^{- 1} \frac{3}{\sqrt{10}} \Rightarrow \tan^{- 1} x + \tan^{- 1} [\frac{\sqrt{1 - {(\frac{y}{\sqrt{1 + y^{2}}})}^{2}}}{\frac{y}{\sqrt{1 + y^{2}}}}] = \tan^{- 1} [\frac{\frac{3}{\sqrt{10}}}{\sqrt{1 - {(\frac{3}{\sqrt{10}})}^{2}}}] \Rightarrow \tan^{- 1} x + \tan^{- 1} (\frac{1}{y}) = \tan^{- 1} (3) \Rightarrow \tan^{- 1} (\frac{x + \frac{1}{y}}{1 - x \times \frac{1}{y}}) = \tan^{- 1} (3) \Rightarrow \frac{x y + 1}{y - x} = 3 \Rightarrow 3 y - 3 x = x y + 1 \Rightarrow 3 x + x y = 3 y - 1 \Rightarrow x (3 + y) = 3 y - 1 \Rightarrow x = \frac{3 y - 1}{3 + y} For, y = 1 \Rightarrow x = \frac{1}{2} For, y = 2 \Rightarrow x = 1 For, y = 3 \Rightarrow x = \frac{4}{3} For, y = 4 \Rightarrow x = \frac{11}{7} For, y = 1 \Rightarrow x = \frac{7}{3} and so on . . . . . . Therefore, only integral solutions are : x = 1 and y = 2$

Page No 3.117:

Question 6:

If sin⁻¹ x − cos⁻¹ x = $\frac{π}{6}$ , then x =
(a) $\frac{1}{2}$

(b) $\frac{\sqrt{3}}{2}$

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

(d) none of these

Answer:

(b) $\frac{\sqrt{3}}{2}$
We know that $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ .

$∴ \sin^{- 1} x - \cos^{- 1} x = \frac{π}{6} \Rightarrow \frac{π}{2} - \cos^{- 1} x - \cos^{- 1} x = \frac{π}{6} \Rightarrow - 2 \cos^{- 1} x = \frac{π}{6} - \frac{π}{2} \Rightarrow - 2 \cos^{- 1} x = - \frac{π}{3} \Rightarrow \cos^{- 1} x = \frac{π}{6} \Rightarrow x = \cos \frac{π}{6} \Rightarrow x = \frac{\sqrt{3}}{2}$

Page No 3.117:

Question 7:

sin $[\cot^{- 1} \{\tan (\cos^{- 1} x)\}]$ is equal to
(a) x

(b) $\sqrt{1 - x^{2}}$

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

(d) none of these

Answer:

(a) x

Let $\cos^{- 1} x = y$

Then,
$\sin [\cot^{- 1} \{\tan (\cos^{- 1} x)\}] = \sin [\cot^{- 1} \{\tan y\}] = \sin [\cot^{- 1} \{\cot (\frac{π}{2} - y)\}] = \sin (\frac{π}{2} - y) = \cos y = x [∵ \cos y = x]$

Page No 3.117:

Question 8:

The number of solutions of the equation
$\tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4}$ is
(a) 2
(b) 3
(c) 1
(d) none of these

Answer:

(a) 2
We know that $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$ .

$∴ \tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{2 x + 3 x}{1 - 2 x \times 3 x}) = \frac{π}{4} \Rightarrow \frac{2 x + 3 x}{1 - 2 x \times 3 x} = \tan \frac{π}{4} \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \Rightarrow 5 x = 1 - 6 x^{2} \Rightarrow 6 x^{2} + 5 x - 1 = 0$

Therefore, there are two solutions.

Page No 3.117:

Question 9:

If α = $\tan^{- 1} (\tan \frac{5 π}{4}) and β = \tan^{- 1} (- \tan \frac{2 π}{3})$ , then
(a) 4 α = 3 β
(b) 3 α = 4 β
(c) α − β = $\frac{7 π}{12}$
(d) none of these

Answer:

(a) 4 α = 3 β

We know that $\tan^{- 1} (\tan x) = x$ .

$∴ α = \tan^{- 1} (\tan \frac{5 π}{4}) = \tan^{- 1} \{\tan (π + \frac{π}{4})\} = \tan^{- 1} (\tan \frac{π}{4}) = \frac{π}{4}$
and
$β = \tan^{- 1} \{- \tan (\frac{2 π}{3})\} = \tan^{- 1} \{- \tan (π - \frac{π}{3})\} = \tan^{- 1} \{\tan (\frac{π}{3})\} = \frac{π}{3}$

$∴ 4 α = π 3 β = π$
∴ $4 α = 3 β$

Page No 3.117:

Question 10:

The number of real solutions of the equation
$\sqrt{1 + \cos 2 x} = \sqrt{2} \sin^{- 1} (\sin x), - π \leq x \leq π$ is
(a) 0
(b) 1
(c) 2
(d) infinite

Answer:

(c) 2

$For, - π \leq x \leq \frac{- π}{2} \sqrt{1 + \cos 2 x} = \sqrt{2} \sin^{- 1} (\sin x) \Rightarrow \sqrt{2} |\cos x| = \sqrt{2} (- π - x) \Rightarrow \sqrt{2} (- \cos x) = \sqrt{2} (- π - x) \Rightarrow \cos x = π + x It does not satisfy for any value of x in the interval (- π, \frac{- π}{2})$

$For, \frac{- π}{2} \leq x \leq \frac{π}{2} \sqrt{1 + \cos 2 x} = \sqrt{2} \sin^{- 1} (\sin x) \Rightarrow \sqrt{2} |\cos x| = \sqrt{2} (x) \Rightarrow \sqrt{2} (\cos x) = \sqrt{2} (x) \Rightarrow \cos x = x It gives one value of x in the interval (\frac{- π}{2}, \frac{π}{2})$

$For, \frac{π}{2} \leq x \leq π \sqrt{1 + \cos 2 x} = \sqrt{2} \sin^{- 1} (\sin x) \Rightarrow \sqrt{2} |\cos x| = \sqrt{2} (- π - x) \Rightarrow \sqrt{2} (- \cos x) = \sqrt{2} (π - x) \Rightarrow \cos x = - π + x It gives one value of x in the interval (\frac{π}{2}, π)$

$∴ \sqrt{1 + \cos 2 x} = \sqrt{2} \sin^{- 1} (\sin x) gives two real solutions in the interval [- π, π]$

Page No 3.117:

Question 11:

If x < 0, y < 0 such that xy = 1, then tan⁻¹ x + tan⁻¹ y equals
(a) $\frac{π}{2}$

(b) $- \frac{π}{2}$

(c) − π

(d) none of these

Answer:

(b) $- \frac{π}{2}$
We know that $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$ .
$x < 0, y < 0$ such that
$x y = 1$

Let x = $-$ a and y = $-$ b, where a and b both are positive.

$∴ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y}) = \tan^{- 1} (\frac{- a - a}{1 - 1}) = \tan^{- 1} (- \infty) = \tan^{- 1} \{\tan (- \frac{π}{2})\} = - \frac{π}{2}$

Page No 3.117:

Question 12:

$If u = \cot^{- 1} \sqrt{\tan θ} - \tan^{- 1} \sqrt{\tan θ} then, \tan (\frac{π}{4} - \frac{u}{2}) =$
(a) $\sqrt{\tan θ}$

(b) $\sqrt{\cot θ}$

(c) tan θ

(d) cot θ

Answer:

(a) $\sqrt{\tan θ}$

Let $y = \sqrt{\tan θ}$
Then,
$u = \cot^{- 1} \sqrt{\tan θ} - \tan^{- 1} \sqrt{\tan θ} \Rightarrow u = \cot^{- 1} y - \tan^{- 1} y \Rightarrow u = \frac{π}{2} - 2 \tan^{- 1} y [∵ \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}] \Rightarrow 2 \tan^{- 1} y = \frac{π}{2} - u \Rightarrow \tan^{- 1} y = \frac{π}{4} - \frac{u}{2} \Rightarrow y = \tan (\frac{π}{4} - \frac{u}{2}) \Rightarrow \sqrt{\tan θ} = \tan (\frac{π}{4} - \frac{u}{2}) [∵ y = \sqrt{\tan θ}]$

Page No 3.117:

Question 13:

$If \cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{θ}{2}, then 4 x^{2} - 12 x y \cos \frac{θ}{2} + 9 y^{2} =$
(a) 36
(b) 36 − 36 cos θ
(c) 18 − 18 cos θ
(d) 18 + 18 cos θ

Answer:

(c) 18 − 18 cosθ

We know
$\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} (x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}})$

$∴ \cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{θ}{2} \Rightarrow \cos^{- 1} (\frac{x}{3} \frac{y}{2} - \sqrt{1 - \frac{x^{2}}{9}} \sqrt{1 - \frac{y^{2}}{4}}) = \frac{θ}{2} \Rightarrow \frac{x y}{6} - \sqrt{\frac{9 - x^{2}}{9}} \sqrt{\frac{4 - y^{2}}{4}} = \cos \frac{θ}{2} \Rightarrow x y - 6 \cos \frac{θ}{2} = \sqrt{9 - x^{2}} \sqrt{4 - y^{2}}$

Squaring both the sides, we get
$x^{2} y^{2} - 12 x y \cos \frac{θ}{2} + 36 \cos^{2} \frac{θ}{2} = (9 - x^{2}) (4 - y^{2}) \Rightarrow x^{2} y^{2} - 12 x y \cos \frac{θ}{2} + 36 \cos^{2} \frac{θ}{2} = 36 - 9 y^{2} - 4 x^{2} + x^{2} y^{2} \Rightarrow 4 x^{2} + 9 y^{2} - 12 x y \cos^{2} \frac{θ}{2} = 36 - 36 \cos^{2} \frac{θ}{2} \Rightarrow 4 x^{2} + 9 y^{2} - 12 x y \cos^{2} \frac{θ}{2} = 36 \{1 - (\frac{\cos θ + 1}{2})\} [∵ \cos 2 x = 2 \cos^{2} x - 1] \Rightarrow 4 x^{2} + 9 y^{2} - 12 x y \cos^{2} \frac{θ}{2} = 18 - 18 \cos θ$

Page No 3.117:

Question 14:

If α = $\tan^{- 1} (\frac{\sqrt{3} x}{2 y - x}), β = \tan^{- 1} (\frac{2 x - y}{\sqrt{3} y}),$ then α − β =
(a) $\frac{π}{6}$

(b) $\frac{π}{3}$

(c) $\frac{π}{2}$

(d) $- \frac{π}{3}$

Answer:

(a) $\frac{π}{6}$

We have
α = $\tan^{- 1} (\frac{\sqrt{3} x}{2 y - x}), β = \tan^{- 1} (\frac{2 x - y}{\sqrt{3} y})$
$Now, α - β = \tan^{- 1} (\frac{\sqrt{3} x}{2 y - x}) - \tan^{- 1} \frac{2 x - y}{\sqrt{3} y} = \tan^{- 1} (\frac{\frac{\sqrt{3} x}{2 y - x} - \frac{2 x - y}{\sqrt{3} y}}{1 + \frac{\sqrt{3} x}{2 y - x} \times \frac{2 x - y}{\sqrt{3} y}}) = \tan^{- 1} (\frac{\frac{3 x y - 4 x y + 2 y^{2} + 2 x^{2} - x y}{\sqrt{3} y (2 y - x)}}{\frac{\sqrt{3} y (2 y - x) + \sqrt{3} x (2 x - y)}{\sqrt{3} y (2 y - x)}}) = \tan^{- 1} (\frac{3 x y - 4 x y + 2 y^{2} + 2 x^{2} - x y}{2 \sqrt{3} y^{2} - \sqrt{3} x y + 2 \sqrt{3} x^{2} - \sqrt{3} x y}) = \tan^{- 1} (\frac{2 y^{2} + 2 x^{2} - 2 x y}{2 \sqrt{3} y^{2} + 2 \sqrt{3} x^{2} - 2 \sqrt{3} x y}) = \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$

Page No 3.117:

Question 15:

Let f (x) = $e^{\cos^{- 1} \{\sin (x + π / 3)\}}$ . Then, f (8π/9) =
(a) e⁵^π^/18
(b) e^13π^/18
(c) e⁻²^π/18
(d) none of these

Answer:

(b) e^13π^/18

Given: $f (x) = e^{\cos^{- 1} \{\sin (x + \frac{π}{3})\}}$

Then,
$f (\frac{8 π}{9}) = e^{\cos^{- 1} \{\sin (\frac{8 π}{9} + \frac{π}{3})\}} = e^{\cos^{- 1} \{\sin (\frac{11 π}{9})\}} = e^{\cos^{- 1} \{\cos (\frac{π}{2} + \frac{13 π}{18})\}} [∵ \cos (\frac{π}{2} + θ) = \sin θ] = e^{\cos^{- 1} \{\cos (\frac{13 π}{18})\}} = e^{\frac{13 π}{18}}$

Page No 3.118:

Question 16:

$\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}$ is equal to
(a) 0
(b) 1/2
(c) − 1
(d) none of these

Answer:

(d) none of these

We know that $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$ .
Now,
$\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11} = \tan^{- 1} (\frac{\frac{1}{11} + \frac{2}{11}}{1 - \frac{1}{11} \frac{2}{11}}) = \tan^{- 1} (\frac{\frac{3}{11}}{\frac{121 - 2}{121}}) = \tan^{- 1} (\frac{\frac{3}{11}}{\frac{119}{121}}) = \tan^{- 1} (\frac{33}{119}) = 0.27$

Page No 3.118:

Question 17:

If $\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = θ,$ then 9x² − 12xy cos θ + 4y² is equal to
(a) 36
(b) −36 sin² θ
(c) 36 sin² θ
(d) 36 cos² θ

Answer:

(c) 36 sin² θ

We know
$\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} [x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}]$

Now,
$\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = θ \Rightarrow \cos^{- 1} [\frac{x}{2} \frac{y}{3} - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{3}}] = θ \Rightarrow \frac{x}{2} \frac{y}{3} - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{3}} = \cos θ \Rightarrow x y - \sqrt{4 - x^{2}} \sqrt{9 - y^{2}} = 6 \cos θ \Rightarrow \sqrt{4 - x^{2}} \sqrt{9 - y^{2}} = x y - 6 \cos θ \Rightarrow (4 - x^{2}) (9 - y^{2}) = x^{2} y^{2} + 36 \cos^{2} θ - 12 x y \cos θ (Squaring both the sides) \Rightarrow 36 - 4 y^{2} - 9 x^{2} + x^{2} y^{2} = x^{2} y^{2} + 36 \cos^{2} θ - 12 x y \cos θ \Rightarrow 36 - 4 y^{2} - 9 x^{2} = 36 \cos^{2} θ - 12 x y \cos θ \Rightarrow 9 x^{2} - 12 x y \cos θ + 4 y^{2} = 36 - 36 \cos^{2} θ \Rightarrow 9 x^{2} - 12 x y \cos θ + 4 y^{2} = 36 \sin^{2} θ$

Page No 3.118:

Question 18:

If tan⁻¹ 3 + tan⁻¹ x = tan⁻¹ 8, then x =
(a) 5
(b) 1/5
(c) 5/14
(d) 14/5

Answer:

(b) $\frac{1}{5}$
We know that $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}$ .
Now,
$\tan^{- 1} 3 + \tan^{- 1} x = \tan^{- 1} 8 \Rightarrow \tan^{- 1} (\frac{3 + x}{1 - 3 x}) = \tan^{- 1} 8 \Rightarrow \frac{3 + x}{1 - 3 x} = 8 \Rightarrow 3 + x = 8 - 24 x \Rightarrow 3 - 8 = - 24 x - x \Rightarrow - 5 = - 25 x \Rightarrow x = \frac{5}{25} = \frac{1}{5}$

Page No 3.118:

Question 19:

The value of $\sin^{- 1} (\cos \frac{33 π}{5})$ is
(a) $\frac{3 π}{5}$

(b) $- \frac{π}{10}$

(c) $\frac{π}{10}$

(d) $\frac{7 π}{5}$

Answer:

(b) $- \frac{π}{10}$

$\sin^{- 1} (\cos \frac{33 π}{5}) = \sin^{- 1} \{\cos (6 π + \frac{3 π}{5})\} = \sin^{- 1} \{\cos (\frac{3 π}{5})\} = \sin^{- 1} \{\sin (\frac{π}{2} - \frac{3 π}{5})\} = \frac{π}{2} - \frac{3 π}{5} = - \frac{π}{10}$

Page No 3.118:

Question 20:

The value of $\cos^{- 1} (\cos \frac{5 π}{3}) + \sin^{- 1} (\sin \frac{5 π}{3})$ is
(a) $\frac{π}{2}$

(b) $\frac{5 π}{3}$

(c) $\frac{10 π}{3}$

(d) 0

Answer:

(d) 0

We have
$\cos^{- 1} (\cos \frac{5 π}{3}) + \sin^{- 1} (\sin \frac{5 π}{3}) = \cos^{- 1} \{\cos (2 π - \frac{π}{3})\} + \sin^{- 1} \{\sin (2 π - \frac{π}{3})\} = \cos^{- 1} \{\cos (\frac{π}{3})\} + \sin^{- 1} \{- \sin (\frac{π}{3})\} = \cos^{- 1} \{\cos (\frac{π}{3})\} - \sin^{- 1} \{\sin (\frac{π}{3})\} = \frac{π}{3} - \frac{π}{3} = 0$

Page No 3.118:

Question 21:

sin $\{2 \cos^{- 1} (\frac{- 3}{5})\}$ is equal to
(a) $\frac{6}{25}$

(b) $\frac{24}{25}$

(c) $\frac{4}{5}$

(d) $- \frac{24}{25}$

Answer:

(d) $- \frac{24}{25}$

Let $\cos^{- 1} (- \frac{3}{5}) = x, 0 \leq x \leq π$
Then, $\cos x = - \frac{3}{5}$

$∴ \sin x = \sqrt{1 - \cos^{2} x} = \sqrt{1 - {(- \frac{3}{5})}^{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$
Now,
$\sin \{2 \cos^{- 1} (- \frac{3}{5})\} = \sin (2 x) = 2 \sin x \cos x = 2 \times \frac{4}{5} \times \frac{- 3}{5} = - \frac{24}{25}$

Page No 3.118:

Question 22:

If θ = sin⁻¹ {sin (−600°)}, then one of the possible values of θ is
(a) $\frac{π}{3}$

(b) $\frac{π}{2}$

(c) $\frac{2 π}{3}$

(d) $- \frac{2 π}{3}$

Answer:

(a) $\frac{π}{3}$

We know
$\sin^{- 1} (\sin x) = x$

Now,

$θ = \sin^{- 1} \{\sin (- 600^{\circ})\} = \sin^{- 1} \{\sin (720^{\circ} - 600^{\circ})\} = \sin^{- 1} \{\sin (120^{\circ})\} = \sin^{- 1} \{\sin (180^{\circ} - 120^{\circ})\} [∵ \sin x = \sin (π - x)] = \sin^{- 1} (\sin 60^{\circ}) = 60^{\circ}$

Page No 3.118:

Question 23:

If 3 $\sin^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}$ is equal to
(a) $\frac{1}{\sqrt{3}}$

(b) $- \frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $- \frac{\sqrt{3}}{4}$

Answer:

(a) $\frac{1}{\sqrt{3}}$

Let $x = \tan y$
Then,
$3 \sin^{- 1} (\frac{2 \tan y}{1 + \tan^{2} y}) - 4 (\frac{1 - \tan^{2} y}{1 + \tan^{2} y}) + 2 \tan^{- 1} (\frac{2 \tan y}{1 - \tan^{2} y}) = \frac{π}{3} \Rightarrow 3 \sin^{- 1} (\sin 2 y) - 4 \cos^{- 1} (\cos 2 y) + 2 \tan^{- 1} (\tan 2 y) = \frac{π}{3} [∵ \sin 2 y = (\frac{2 \tan y}{1 + \tan^{2} y}), \cos 2 y = (\frac{1 - \tan^{2} y}{1 + \tan^{2} y}) and \tan 2 y = (\frac{2 \tan y}{1 - \tan^{2} y})] \Rightarrow 3 \times 2 y - 4 \times 2 y + 2 \times 2 y = \frac{π}{3} \Rightarrow 6 y - 8 y + 4 y = \frac{π}{3} \Rightarrow 2 y = \frac{π}{3} \Rightarrow y = \frac{π}{6} \Rightarrow \tan^{- 1} x = \frac{π}{6} [∵ \tan^{- 1} x = y] \Rightarrow x = \tan \frac{π}{6} \Rightarrow x = \frac{1}{\sqrt{3}}$

Page No 3.118:

Question 24:

If 4 cos⁻¹ x + sin⁻¹x = π, then the value of x is
(a) $\frac{3}{2}$

(b) $\frac{1}{\sqrt{2}}$

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{2}{\sqrt{3}}$

Answer:

(c) $\frac{\sqrt{3}}{2}$
We know that $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ .

$4 \cos^{- 1} x + \sin^{- 1} x = π \Rightarrow 4 \cos^{- 1} x + \frac{π}{2} - \cos^{- 1} x = π \Rightarrow 3 \cos^{- 1} x = π - \frac{π}{2} \Rightarrow 3 \cos^{- 1} x = \frac{π}{2} \Rightarrow \cos^{- 1} x = \frac{π}{6} \Rightarrow x = \cos \frac{π}{6} \Rightarrow x = \frac{\sqrt{3}}{2}$

Page No 3.118:

Question 25:

It $\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}$ (−7), then the value of x is
(a) 0
(b) −2
(c) 1
(d) 2

Answer:

(d) 2

We know that $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$ .

$∴ \tan^{- 1} (\frac{x + 1}{x - 1}) + \tan^{- 1} (\frac{x - 1}{x}) = \tan^{- 1} (- 7) \Rightarrow \tan^{- 1} (\frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x}}{1 - \frac{x + 1}{x - 1} \times \frac{x - 1}{x}}) = \tan^{- 1} (- 7) \Rightarrow \tan^{- 1} (\frac{\frac{x^{2} + x + x^{2} - 2 x + 1}{x (x - 1)}}{\frac{x^{2} - x - x^{2} + 1}{x (x - 1)}}) = \tan^{- 1} (- 7) \Rightarrow \tan^{- 1} (\frac{2 x^{2} - x + 1}{- x + 1}) = \tan^{- 1} (- 7)$

So, we get

$\frac{2 x^{2} - x + 1}{- x + 1} = - 7 \Rightarrow 2 x^{2} - x + 1 = 7 x - 7 \Rightarrow 2 x^{2} - 8 x + 8 = 0 \Rightarrow x^{2} - 4 x + 4 = 0 \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x = 2$

Page No 3.118:

Question 26:

If $\cos^{- 1} x > \sin^{- 1} x$ , then

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

Answer:

$\cos^{- 1} x > \sin^{- 1} x \Rightarrow \cos^{- 1} x > \frac{π}{2} - \cos^{- 1} x \Rightarrow 2 \cos^{- 1} x > \frac{π}{2} \Rightarrow \cos^{- 1} x > \frac{π}{4} \Rightarrow x > \cos \frac{π}{4} \Rightarrow x > \frac{1}{\sqrt{2}}$
We know that the maximum value of cosine fuction is 1.
$∴ \frac{1}{\sqrt{2}} < x \leq 1$
Hence, the correct answer is option(a).

Page No 3.118:

Question 27:

In a ∆ ABC, if C is a right angle, then
$\tan^{- 1} (\frac{a}{b + c}) + \tan^{- 1} (\frac{b}{c + a}) =$

(a) $\frac{π}{3}$

(b) $\frac{π}{4}$

(c) $\frac{5 π}{2}$

(d) $\frac{π}{6}$

Answer:

(b) $\frac{π}{4}$

We know
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$

$∴ \tan^{- 1} (\frac{a}{b + c}) + \tan^{- 1} (\frac{b}{c + a}) = \tan^{- 1} (\frac{\frac{a}{b + c} + \frac{b}{c + a}}{1 - \frac{a}{b + c} \times \frac{b}{c + a}}) = \tan^{- 1} (\frac{\frac{a c + a^{2} + b^{2} + b c}{(b + c) (c + a)}}{\frac{a c + c^{2} + b c}{(b + c) (c + a)}})$

$= \tan^{- 1} (\frac{a c + c^{2} + b c}{a c + c^{2} + b c}) [∵ a^{2} + b^{2} = c^{2}] = \tan^{- 1} (1) = \tan^{- 1} (\tan \frac{π}{4}) = \frac{π}{4}$

Page No 3.118:

Question 28:

The value of sin $(\frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8})$ is
(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{\sqrt{3}}$

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

(d) $\frac{1}{3 \sqrt{3}}$

Answer:

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

Let $\sin^{- 1} \frac{\sqrt{63}}{8} = y$
Then,
$\sin y = \frac{\sqrt{63}}{8} \cos y = \sqrt{1 - \sin^{2} y} = \sqrt{1 - \frac{63}{64}} = \frac{1}{8}$
Now, we have

$\sin (\frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8}) = \sin (\frac{1}{4} y) = \sqrt{\frac{1 - \cos \frac{y}{2}}{2}} [∵ \cos 2 x = 1 - 2 \sin^{2} x] = \sqrt{\frac{1 - \sqrt{\frac{1 + \cos y}{2}}}{2}} [∵ \cos 2 x = 2 \cos^{2} x - 1] = \sqrt{\frac{1 - \sqrt{\frac{1 + \frac{1}{8}}{2}}}{2}} = \sqrt{\frac{1 - \sqrt{\frac{9}{16}}}{2}} = \sqrt{\frac{1 - \frac{3}{4}}{2}} = \sqrt{\frac{1}{8}} = \frac{1}{2 \sqrt{2}}$

Page No 3.119:

Question 29:

$\cot (\frac{π}{4} - 2 \cot^{- 1} 3) =$
(a) 7
(b) 6
(c) 5
(d) none of these

Answer:

(a) 7

Let $2 \cot^{- 1} 3 = y$
Then, $\cot \frac{y}{2} = 3$

$\cot (\frac{π}{4} - 2 \cot^{- 1} 3) = \cot (\frac{π}{4} - y) = \frac{\cot \frac{π}{4} \cot y + 1}{\cot y - \cot \frac{π}{4}} = \frac{\cot y + 1}{\cot y - 1} = \frac{\frac{\cot^{2} \frac{y}{2} - 1}{2 \cot \frac{y}{2}} + 1}{\frac{\cot^{2} \frac{y}{2} - 1}{2 \cot \frac{y}{2}} - 1} = \frac{\cot^{2} \frac{y}{2} + 2 \cot \frac{y}{2} - 1}{\cot^{2} \frac{y}{2} - 2 \cot \frac{y}{2} - 1} = \frac{9 + 6 - 1}{9 - 6 - 1} = 7$

Page No 3.119:

Question 30:

If tan⁻¹ (cot θ) = 2 θ, then θ =

(a) $\pm \frac{π}{3}$

(b) $\pm \frac{π}{4}$

(c) $\pm \frac{π}{6}$

(d) none of these

Answer:

(c) $\pm \frac{π}{6}$

$We have, \tan^{- 1} (c o t θ) = 2 θ \Rightarrow \tan 2 θ = c o t θ \Rightarrow \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{1}{\tan θ} \Rightarrow 2 \tan^{2} θ = 1 - \tan^{2} θ \Rightarrow 3 \tan^{2} θ = 1 \Rightarrow \tan^{2} θ = \frac{1}{3} \Rightarrow \tan θ = \pm \frac{1}{\sqrt{3}} ∴ θ = \pm \frac{π}{6}$

Page No 3.119:

Question 31:

If $\sin^{- 1} (\frac{2 a}{1 - a^{2}}) + \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), where a, x \in (0, 1)$ , then, the value of x is

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

Answer:

$\sin^{- 1} (\frac{2 a}{1 - a^{2}}) + \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) \Rightarrow 2 \tan^{- 1} a + 2 \tan^{- 1} a = 2 \tan^{- 1} x \Rightarrow 4 \tan^{- 1} a = 2 \tan^{- 1} x \Rightarrow 2 \tan^{- 1} a = \tan^{- 1} x \Rightarrow \tan^{- 1} (\frac{2 a}{1 - a^{2}}) = \tan^{- 1} x \Rightarrow x = \frac{2 a}{1 - a^{2}}$
Hence, the correct answer is option(d).

Page No 3.119:

Question 32:

The value of $\sin (2 (\tan^{- 1} 0.75))$ is equal to
(a) 0.75
(b) 1.5
(c) 0.96
(d) $\sin^{- 1} 1.5$

Answer:

$\sin (2 (\tan^{- 1} 0.75)) = \sin (2 \tan^{- 1} 0.75) = \sin (\sin^{- 1} \frac{2 \times 0.75}{1 + {(0.75)}^{2}}) = \sin (\sin^{- 1} 0.96) = 0.96$
Hence, the correct answer is option (c).

Page No 3.119:

Question 33:

If x > 1, then $2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}})$ is equal to
(a) $4 \tan^{- 1} x$
(b) 0
(c) $\frac{π}{2}$
(d) $π$

Answer:

$2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 \tan^{- 1} x + 2 \tan^{- 1} x [∵ \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 \tan^{- 1} x] = 4 \tan^{- 1} x$
Hence, the correct answer is option (a)

Page No 3.119:

Question 34:

The domain of $\cos^{- 1} (x^{2} - 4)$ is
(a) [3, 5]
(b) [−1, 1]
(c) $[- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$
(d) $[- \sqrt{5}, - \sqrt{3}] \cap [\sqrt{3}, \sqrt{5}]$

Answer:

The domain of $\cos^{- 1} (x)$ is [−1, 1]
$∴ - 1 \leq x^{2} - 4 \leq 1 \Rightarrow - 1 + 4 \leq x^{2} - 4 + 4 \leq 1 + 4 \Rightarrow 3 \leq x^{2} \leq 5 \Rightarrow \pm \sqrt{3} \leq x \leq \pm \sqrt{5} \Rightarrow x \in [- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$
Hence, the correct answer is option (c).

Page No 3.119:

Question 35:

The value of $\tan (\cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4})$
(a) $\frac{19}{8}$
(b) $\frac{8}{19}$
(c) $\frac{19}{12}$
(d) $\frac{3}{4}$

Answer:

$\tan (\cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4}) = \tan (\tan^{- 1} \frac{\sqrt{1 - \frac{9}{25}}}{\frac{3}{5}} + \tan^{- 1} \frac{1}{4}) = \tan (\tan^{- 1} \frac{\frac{4}{5}}{\frac{3}{5}} + \tan^{- 1} \frac{1}{4}) = \tan (\tan^{- 1} \frac{4}{3} + \tan^{- 1} \frac{1}{4}) = \tan (\tan^{- 1} \frac{\frac{4}{3} + \frac{1}{4}}{1 - \frac{1}{3}}) = \frac{\frac{16 + 3}{12}}{\frac{2}{3}} = \frac{19}{8}$
Hence, the correct answer is option (a).

Page No 3.119:

Question 36:

$If α \leq 2 \sin^{- 1} x + \cos^{- 1} x \leq β, then (a) α = - \frac{π}{2}, β = \frac{π}{2} (b) α = 0, β = π (c) α = - \frac{π}{2}, β = \frac{3 π}{2} (d) α = 0, β = 2 π$

Answer:

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

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

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

Now,

$- \frac{π}{2} \leq \sin^{- 1} x \leq \frac{π}{2}$

$\Rightarrow - \frac{π}{2} + \frac{π}{2} \leq \sin^{- 1} x + \frac{π}{2} \leq \frac{π}{2} + \frac{π}{2}$

$\Rightarrow 0 \leq \sin^{- 1} x + \frac{π}{2} \leq π$

$∴ 0 \leq 2 \sin^{- 1} x + \cos^{- 1} x \leq π$

Comparing with $α \leq 2 \sin^{- 1} x + \cos^{- 1} x \leq β$ , we get

$α = 0, β = π$

Hence, the correct answer is option (b).

Page No 3.119:

Question 37:

The value of sin (2sin^-1(.6)) is
(a) 0.48 (b) 0.96 (c) 1.2 (d) sin 1.2

Answer:

$\sin (2 \sin^{- 1} 0.6)$

$= \sin [\sin^{- 1} (2 \times 0.6 \times \sqrt{1 - {(0.6)}^{2}})] [2 \sin^{- 1} x = \sin^{- 1} (2 x \sqrt{1 - x^{2}})]$

$= \sin [\sin^{- 1} (2 \times 0.6 \times \sqrt{1 - 0.36})]$

$= \sin [\sin^{- 1} (2 \times 0.6 \times \sqrt{0.64})]$

$= \sin [\sin^{- 1} (2 \times 0.6 \times 0.8)]$

$= \sin [\sin^{- 1} (0.96)]$

$= 0.96 [\sin (\sin^{- 1} x) = x, \forall x \in [- 1, 1]]$

Hence, the correct answer is option (b).

Page No 3.119:

Question 38:

The value of cot (sin⁻¹x) is
$(a) \frac{\sqrt{1 + x^{2}}}{x} (b) \frac{x}{\sqrt{1 + x^{2}}} (c) \frac{1}{x} (d) \frac{\sqrt{1 - x^{2}}}{x}$

Answer:

We know

$\sin^{- 1} x = \cot^{- 1} \frac{\sqrt{1 - x^{2}}}{x} ∴ \cot (\sin^{- 1} x) = \cot (\cot^{- 1} \frac{\sqrt{1 - x^{2}}}{x}) \Rightarrow \cot (\sin^{- 1} x) = \frac{\sqrt{1 - x^{2}}}{x} [\cot (\cot^{- 1} x) = x, \forall x \in R]$

Thus, the value of cot(sin⁻¹x) is $\frac{\sqrt{1 - x^{2}}}{x}$ .

Hence, the correct answer is option (d).

Page No 3.119:

Question 39:

If tan⁻¹x = $\frac{x}{10}$ for some x ∊ R, then the value of cot⁻¹ x is
$(a) \frac{π}{5} (b) \frac{2 π}{5} (c) \frac{3 π}{5} (d) \frac{4 π}{5}$

Answer:

Disclaimer: The solution has been provided for the following question.

If tan⁻¹x = $\frac{π}{10}$ for some x ∊ R, then the value of cot⁻¹ x is
$(a) \frac{π}{5} (b) \frac{2 π}{5} (c) \frac{3 π}{5} (d) \frac{4 π}{5}$

Solution:

We know

$\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}, \forall x \in R \Rightarrow \frac{π}{10} + \cot^{- 1} x = \frac{π}{2} (\tan^{- 1} x = \frac{π}{10}) \Rightarrow \cot^{- 1} x = \frac{π}{2} - \frac{π}{10} \Rightarrow \cot^{- 1} x = \frac{4 π}{10} = \frac{2 π}{5}$

Hence, the correct answer is option (b).

Page No 3.119:

Question 40:

One branch of cos^-1 other than the principal value branch corresponds to
$(a) [\frac{π}{2}, \frac{3 π}{2}] (b) [π, 2 π] - \{\frac{3 π}{2}\} (c) (0, π) (d) [2 π, 3 π]$

Answer:

The domain of the function $f (x) = \cos^{- 1} x$ is $[- 1, 1]$ . The range of $\cos^{- 1} x$ in one of the intervals $. . ., [- π, 0], [0, π], [π, 2 π], [2 π, 3 π], . . .$ is one-one and onto with the range $[- 1, 1]$ .

Thus, one branch of cos⁻¹x other than the principal value branch corresponds to $[2 π, 3 π]$ .

Hence, the correct answer is option (d).

Page No 3.120:

Question 41:

The principal value branch of sec^-1 is
$(a) [- \frac{π}{2}, \frac{π}{2}] - \{0\} (b) [0, π] - \{\frac{π}{2}\} (c) (0, π) (d) (- \frac{π}{2}, \frac{π}{2})$

Answer:

The principal value branch of sec⁻¹x is $[0, π] - \{\frac{π}{2}\}$ .

Hence, the correct answer is option (b).

Page No 3.120:

Question 42:

Which of the following corresponds to the principal value branch of tan^-1?
(a) $(- \frac{π}{2}, \frac{π}{2}) (b) [- \frac{π}{2}, \frac{π}{2}] (c) (- \frac{π}{2}, \frac{π}{2}) - \{0\} (d) (0, π)$

Answer:

The principal value branch of tan⁻¹x is $(- \frac{π}{2}, \frac{π}{2})$ .

Hence, the correct answer is option (a).

Page No 3.120:

Question 43:

Which of the following is the principal value branch of cosec^-1 ?
(a) $(- \frac{π}{2}, \frac{π}{2}) (b) [0, π] - \{\frac{π}{2}\} (c) [- \frac{π}{2}, \frac{π}{2}] (d) [- \frac{π}{2}, \frac{π}{2}] - \{0\}$

Answer:

The principal value branch of cosec⁻¹x is $[- \frac{π}{2}, \frac{π}{2}] - \{0\}$ .

Hence, the correct answer is option (d).

Page No 3.120:

Question 44:

The value of the expression tan $(\frac{1}{2} \cos^{- 1} \frac{2}{\sqrt{5}})$ is
(a) $2 + \sqrt{5} (b) \sqrt{5} - 2 (c) \frac{\sqrt{5} + 2}{2} (d) 5 + \sqrt{2}$

Answer:

Let $\cos^{- 1} \frac{2}{\sqrt{5}} = θ$ . Then,
$\cos θ = \frac{2}{\sqrt{5}}$

Now,

$\tan (\frac{1}{2} \cos^{- 1} \frac{2}{\sqrt{5}}) = \tan (\frac{θ}{2}) = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}}$
$= \sqrt{\frac{1 - \frac{2}{\sqrt{5}}}{1 + \frac{2}{\sqrt{5}}}} = \sqrt{\frac{\sqrt{5} - 2}{\sqrt{5} + 2}} = \sqrt{\frac{{(\sqrt{5} - 2)}^{2}}{(\sqrt{5} + 2) (\sqrt{5} - 2)}}$
$= \sqrt{\frac{{(\sqrt{5} - 2)}^{2}}{5 - 4}} = \sqrt{5} - 2$

Thus, the value of the given expression is $\sqrt{5} - 2$ .

Hence, the correct answer is option (b).

Page No 3.120:

Question 45:

If 3 tan^-1x + cot^-1x = π, then x equals
(a) 0 (b) 1 (c) -1 (d) $\frac{1}{2}$

Answer:

$3 \tan^{- 1} x + \cot^{- 1} x = π \Rightarrow 2 \tan^{- 1} x + \tan^{- 1} x + \cot^{- 1} x = π \Rightarrow 2 \tan^{- 1} x + \frac{π}{2} = π [\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}]$
$\Rightarrow 2 \tan^{- 1} x = π - \frac{π}{2} = \frac{π}{2} \Rightarrow \tan^{- 1} x = \frac{π}{4} \Rightarrow x = \tan \frac{π}{4} = 1$

Thus, the value of x is 1.

Hence, the correct answer is option (b).

Page No 3.120:

Question 46:

The value of sin^-1 $(\cos \frac{33 π}{5})$ is
(a) $\frac{3 π}{5} (b) - \frac{7 π}{5} (c) \frac{π}{10} (d) - \frac{π}{10}$

Answer:

$\sin^{- 1} (\cos \frac{33 π}{5}) = \sin^{- 1} [\cos (6 π + \frac{3 π}{5})] = \sin^{- 1} (\cos \frac{3 π}{5})$
$= \sin^{- 1} [\sin (\frac{π}{2} - \frac{3 π}{5})] = \sin^{- 1} [\sin (- \frac{π}{10})] = - \frac{π}{10}$

Thus, the value of $\sin^{- 1} (\cos \frac{33 π}{5})$ is $- \frac{π}{10}$ .

Hence, the correct answer is option (d).

Page No 3.120:

Question 47:

tan ^–13 + tan ^–1 $λ$ = tan^–1 $(\frac{3 + λ}{1 - 3 λ})$ is valid for what value of $λ$
(a) $λ = (- \frac{1}{3}, \frac{1}{3})$
(b) $λ > \frac{1}{3}$
(c) $λ < \frac{1}{3}$
(d) all real values of $λ$

Answer:

Given: tan ^–13 + tan ^–1 $λ$ = tan^–1 $(\frac{3 + λ}{1 - 3 λ})$
We know,
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y}) i f x y < 1 ∴ 3 λ < 1 \Rightarrow λ < \frac{1}{3}$
Hence, the correct answer is option C.

Page No 3.120:

Question 48:

The value of tan^{–1 $(\tan \frac{7 π}{6})$}is
(a) $\frac{π}{6} (b) \frac{π}{2} (c) \frac{π}{3} (d) \frac{7 π}{6}$

Answer:

We know,
$\tan^{- 1} (\tan x) = x; i f x t (\frac{- π}{2}, \frac{π}{2}) ∴ \tan^{- 1} (\tan \frac{7 π}{6}) = \tan^{- 1} (\tan (π + \frac{π}{6})) = \tan^{- 1} (\tan \frac{π}{6}) (∵ \tan (π + θ) = \tan θ) = \frac{π}{6} (∵ \frac{- π}{2} < \frac{π}{6} < \frac{π}{2})$
Thus, the value of $\tan^{- 1} (\tan \frac{7 π}{6}) is \frac{π}{6}$
Hence, the correct answer is option A.

Page No 3.120:

Question 1:

The value of sec²(tan^-12) + cosec² (cot^-1 3) is ____________________.

Answer:

We know

$\tan^{- 1} x = \sec^{- 1} \sqrt{1 + x^{2}}$ and $\cot^{- 1} x = {cosec}^{- 1} \sqrt{1 + x^{2}}$

So,

$\tan^{- 1} 2 = \sec^{- 1} \sqrt{1 + 2^{2}} = \sec^{- 1} \sqrt{5}$

$\cot^{- 1} 3 = {cosec}^{- 1} \sqrt{1 + 3^{2}} = {cosec}^{- 1} \sqrt{10}$

$∴ \sec^{2} (\tan^{- 1} 2) + {cosec}^{2} (\cot^{- 1} 3)$

$= \sec^{2} (\sec^{- 1} \sqrt{5}) + {cosec}^{2} ({cosec}^{- 1} \sqrt{10})$

$= {[\sec (\sec^{- 1} \sqrt{5})]}^{2} + {[cosec ({cosec}^{- 1} \sqrt{10})]}^{2}$

$= {(\sqrt{5})}^{2} + {(\sqrt{10})}^{2}$

$= 15$

Thus, the value of $\sec^{2} (\tan^{- 1} 2) + {cosec}^{2} (\cot^{- 1} 3)$ is 15.

The value of sec²(tan⁻¹2) + cosec²(cot⁻¹ 3) is ____15____.

Page No 3.120:

Question 2:

If sin^-1x - cos^-1x = $\frac{π}{6}$ , then x = _________________________.

Answer:

Given: $\sin^{- 1} x - \cos^{- 1} x = \frac{π}{6}$ .....(1)

We know

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

Adding (1) and (2), we get

$2 \sin^{- 1} x = \frac{π}{6} + \frac{π}{2} \Rightarrow 2 \sin^{- 1} x = \frac{4 π}{6} \Rightarrow \sin^{- 1} x = \frac{π}{3}$
$\Rightarrow x = \sin \frac{π}{3} = \frac{\sqrt{3}}{2}$

If sin⁻¹x − cos⁻¹x = $\frac{π}{6}$ , then x = $\frac{\sqrt{3}}{2}$ .

Page No 3.120:

Question 3:

The range of sin^-1x + cos^-1x + tan^-1x is _______________________.

Answer:

Domain of the given function = $[- 1, 1] \cap R$ = $[- 1, 1]$

Now,

For $- 1 \leq x \leq 1$ ,

$\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ and $- \frac{π}{4} \leq \tan^{- 1} x \leq \frac{π}{4}$

$∴ \frac{π}{2} - \frac{π}{4} \leq \sin^{- 1} x + \cos^{- 1} x + \tan^{- 1} x \leq \frac{π}{2} + \frac{π}{4}$

$\Rightarrow \frac{π}{4} \leq \sin^{- 1} x + \cos^{- 1} x + \tan^{- 1} x \leq \frac{3 π}{4}$

Thus, the range of the given function is $[\frac{π}{4}, \frac{3 π}{4}]$ .

The range of sin⁻¹x + cos⁻¹x + tan⁻¹x is $[\frac{π}{4}, \frac{3 π}{4}]$ .

Page No 3.120:

Question 4:

If sin^-1x = $\frac{π}{5}$ for some x ∊ (-1, 1), then the value of cos^-1 x is ____________________.

Answer:

Given: $\sin^{- 1} x = \frac{π}{5}, \forall x \in (- 1, 1)$

We know

$\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} \Rightarrow \frac{π}{5} + \cos^{- 1} x = \frac{π}{2} \Rightarrow \cos^{- 1} x = \frac{π}{2} - \frac{π}{5} = \frac{3 π}{10}$

Thus, the value of cos⁻¹x is $\frac{3 π}{10}$ .

If sin⁻¹x = $\frac{π}{5}$ for some x ∈ (−1, 1), then the value of cos⁻¹ x is $\frac{3 π}{10}$ .

Page No 3.121:

Question 5:

If x < 0, then tan^-1x + tan^-1 $\frac{1}{x}$ is equal to ____________________.

Answer:

We know

$\tan^{- 1} \frac{1}{x} = \{\begin{cases} \cot^{- 1} x, & for x > 0 \\ - π + \cot^{- 1} x, & for x < 0 \end{cases}$

$∴ \tan^{- 1} x + \tan^{- 1} \frac{1}{x} = \tan^{- 1} x + \cot^{- 1} x - π (x < 0) = \frac{π}{2} - π (\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}, \forall x \in R) = - \frac{π}{2}$

If x < 0, then tan⁻¹x + tan⁻¹ $\frac{1}{x}$ is equal to $- \frac{π}{2}$ .

Page No 3.121:

Question 6:

The value of tan^-12 + tan^-13 is ___________________.

Answer:

We know
$\tan^{- 1} x + \tan^{- 1} y = π + \tan^{- 1} (\frac{x + y}{1 - x y}), if x y > 1$
$∴ \tan^{- 1} 2 + \tan^{- 1} 3 = π + \tan^{- 1} (\frac{2 + 3}{1 - 2 \times 3}) = π + \tan^{- 1} (- 1)$
$= π - \frac{π}{4} = \frac{3 π}{4}$

The value of tan⁻¹2 + tan⁻¹3 is $\frac{3 π}{4}$ .

Page No 3.121:

Question 7:

If tan^-1 $- \frac{1}{\sqrt{3}}$ + cot^-1 x = $\frac{x}{2}$ , then x = __________________.

Answer:

Disclaimer: The solution is provided for the following question.

If tan⁻¹ $- \frac{1}{\sqrt{3}}$ + cot⁻¹ x = $\frac{π}{2}$ , then x = __________________.

Solution:

We know

$\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}$ , for all x ∈ R

$∴ \tan^{- 1} (- \frac{1}{\sqrt{3}}) + \cot^{- 1} (- \frac{1}{\sqrt{3}}) = \frac{π}{2}$ .....(1)

It is given that,

$\tan^{- 1} (- \frac{1}{\sqrt{3}}) + \cot^{- 1} x = \frac{π}{2}$ .....(2)

From (1) and (2), we get

$x = - \frac{1}{\sqrt{3}}$

If tan⁻¹ $- \frac{1}{\sqrt{3}}$ + cot⁻¹ x = $\frac{π}{2}$ , then x = $- \frac{1}{\sqrt{3}}$ .

Page No 3.121:

Question 8:

If tan^-1 x-tan^-1 y = $\frac{π}{4}$ , then x - y - xy = ____________________.

Answer:

$\tan^{- 1} x - \tan^{- 1} y = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{x - y}{1 + x y}) = \frac{π}{4} \Rightarrow \frac{x - y}{1 + x y} = \tan \frac{π}{4}$
$\Rightarrow \frac{x - y}{1 + x y} = 1 \Rightarrow x - y = 1 + x y \Rightarrow x - y - x y = 1$

If tan⁻¹ x − tan⁻¹ y = $\frac{π}{4}$ , then x − y − xy = __1__.

Page No 3.121:

Question 9:

The value of cot (tan^-1x + cot^-1x) for all x ∊ R, is ____________________

Answer:

We know

$\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}$ , for all x ∈ R

$∴ \cot (\tan^{- 1} x + \cot^{- 1} x) = \cot \frac{π}{2}$

$\Rightarrow \cot (\tan^{- 1} x + \cot^{- 1} x) = 0$

The value of cot(tan⁻¹x + cot⁻¹x) for all x ∈ R, is __0__.

Page No 3.121:

Question 10:

If cos^-1x + cos^-1 y = $\frac{π}{3}$ , then sin^-1 x + sin^-1 y =____________________.

Answer:

We know

$\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ , for all x ∈ R .....(1)

Also, $\sin^{- 1} y + \cos^{- 1} y = \frac{π}{2}$ , for all y ∈ R .....(2)

Adding (1) and (2), we get

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

$\Rightarrow \sin^{- 1} x + \sin^{- 1} y + \cos^{- 1} x + \cos^{- 1} y = π$

$\Rightarrow \sin^{- 1} x + \sin^{- 1} y + \frac{π}{3} = π$ (Given)

$\Rightarrow \sin^{- 1} x + \sin^{- 1} y = π - \frac{π}{3} = \frac{2 π}{3}$

If cos⁻¹x + cos⁻¹y = $\frac{π}{3}$ , then sin⁻¹x + sin⁻¹y = $\frac{2 π}{3}$ .

Page No 3.121:

Question 11:

If x > 0, y > 0, xy > 1, then tan^-1x + tan^-1y = _____________________.

Answer:

We know

$\tan^{- 1} x + \tan^{- 1} y = π + \tan^{- 1} (\frac{x + y}{1 - x y})$ , if x > 0, y > 0 and xy > 1

If x > 0, y > 0, xy > 1, then tan⁻¹x + tan⁻¹y = $π + \tan^{- 1} (\frac{x + y}{1 - x y})$ .

Page No 3.121:

Question 12:

If 3 sin^-1x = π-cos^-1x, then x = __________________.

Answer:

$3 \sin^{- 1} x = π - \cos^{- 1} x \Rightarrow 3 (\frac{π}{2} - \cos^{- 1} x) = π - \cos^{- 1} x (\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}) \Rightarrow \frac{3 π}{2} - 3 \cos^{- 1} x = π - \cos^{- 1} x$
$\Rightarrow 2 \cos^{- 1} x = \frac{3 π}{2} - π = \frac{π}{2} \Rightarrow \cos^{- 1} x = \frac{π}{4} \Rightarrow x = \cos \frac{π}{4} = \frac{1}{\sqrt{2}}$

If 3sin⁻¹x = $π$ − cos⁻¹x, then x = $\frac{1}{\sqrt{2}}$ .

Page No 3.121:

Question 13:

If tan^-1x + tan^-1 y = $\frac{5 π}{6}$ , then cot^-1x + cot^-1y = _________________.

Answer:

We know

$\tan^{- 1} a + \cot^{- 1} a = \frac{π}{2}$ , for all a ∈ R .....(1)

Now,

$\tan^{- 1} x + \tan^{- 1} y = \frac{5 π}{6}$ (Given)

$\Rightarrow \frac{π}{2} - \cot^{- 1} x + \frac{π}{2} - \cot^{- 1} y = \frac{5 π}{6}$ [Using (1)]

$\Rightarrow \cot^{- 1} x + \cot^{- 1} y = π - \frac{5 π}{6}$

$\Rightarrow \cot^{- 1} x + \cot^{- 1} y = \frac{π}{6}$

If tan⁻¹x + tan⁻¹ y = $\frac{5 π}{6}$ , then cot⁻¹x + cot⁻¹y = $\frac{π}{6}$ .

Page No 3.121:

Question 14:

If tan^-1x - cot^-1x = tan^-1 $\sqrt{3}$ , then x = _______________________.

Answer:

$\tan^{- 1} x - \cot^{- 1} x = \tan^{- 1} \sqrt{3} \Rightarrow \tan^{- 1} x - (\frac{π}{2} - \tan^{- 1} x) = \frac{π}{3} (\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}) \Rightarrow 2 \tan^{- 1} x = \frac{π}{2} + \frac{π}{3} \Rightarrow \tan^{- 1} x = \frac{π}{4} + \frac{π}{6}$
$\Rightarrow x = \tan (\frac{π}{4} + \frac{π}{6}) \Rightarrow x = \frac{\tan \frac{π}{4} + \tan \frac{π}{6}}{1 - \tan \frac{π}{4} \times \tan \frac{π}{6}} \Rightarrow x = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$
$\Rightarrow x = \frac{{(\sqrt{3} + 1)}^{2}}{(\sqrt{3} - 1) (\sqrt{3} + 1)} \Rightarrow x = \frac{4 + 2 \sqrt{3}}{2} \Rightarrow x = 2 + \sqrt{3}$

If tan⁻¹x − cot⁻¹x = tan⁻¹ $\sqrt{3}$ , then x = $2 + \sqrt{3}$ .

Page No 3.121:

Question 15:

If sin^-1x + sin^-1y + sin^-1z = $- \frac{3 π}{2}$ , then xyz = __________________.

Answer:

We know

$- \frac{π}{2} \leq \sin^{- 1} a \leq \frac{π}{2}$ , for all a ∈ [−1, 1]

So, the minimum value of sin⁻¹a is $- \frac{π}{2}$ .

Now,

$\sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = - \frac{3 π}{2}$ (Given)

This is possible if

$\sin^{- 1} x = - \frac{π}{2}, \sin^{- 1} y = - \frac{π}{2}$ and $\sin^{- 1} z = - \frac{π}{2}$

⇒ x = −1, y = −1 and z = −1

∴ xyz = (−1) × (−1) × (−1) = −1

If sin⁻¹x + sin⁻¹y + sin⁻¹z = $- \frac{3 π}{2}$ , then xyz = ___−1___.

Page No 3.121:

Question 16:

The value of cos^-1 $\{\sin (\cos^{- 1} \frac{1}{2})\}$ is ________________________.

Answer:

$\cos^{- 1} [\sin (\cos^{- 1} \frac{1}{2})] = \cos^{- 1} [\sin (\frac{π}{3})] (\cos \frac{π}{3} = \frac{1}{2} \Rightarrow \cos^{- 1} \frac{1}{2} = \frac{π}{3}) = \cos^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{6} (\cos \frac{π}{6} = \frac{\sqrt{3}}{2} \Rightarrow \cos^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{6})$

Thus, the value of $\cos^{- 1} [\sin (\cos^{- 1} \frac{1}{2})]$ is $\frac{π}{6}$ .

The value of $\cos^{- 1} \{\sin (\cos^{- 1} \frac{1}{2})\}$ is $\frac{π}{6}$ .

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Question 17:

The value of tan $[\cos^{- 1} \{\sin (\cot^{- 1} 1)\}]$ is ___________________.

Answer:

$\tan \{\cos^{- 1} [\sin (\cot^{- 1} 1)]\} = \tan \{\cos^{- 1} [\sin (\frac{π}{4})]\} (\cot \frac{π}{4} = 1 \Rightarrow \cot^{- 1} 1 = \frac{π}{4}) = \tan \{\cos^{- 1} (\frac{1}{\sqrt{2}})\}$
$= \tan \frac{π}{4} (\cos \frac{π}{4} = \frac{1}{\sqrt{2}} \Rightarrow \cos^{- 1} \frac{1}{\sqrt{2}} = \frac{π}{4}) = 1$

Thus, the value of $\tan \{\cos^{- 1} [\sin (\cot^{- 1} 1)]\}$ is 1.

The value of $\tan [\cos^{- 1} \{\sin (\cot^{- 1} 1)\}]$ is __1__.

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Question 18:

The value of tan²(sec^-13) + cot² (cosec^-14) is _________________.

Answer:

$\tan^{2} (\sec^{- 1} 3) + \cot^{2} ({cosec}^{- 1} 4) = \sec^{2} (\sec^{- 1} 3) - 1 + {cosec}^{2} ({cosec}^{- 1} 4) - 1 (1 + \tan^{2} θ = \sec^{2} θ and 1 + \cot^{2} θ = {cosec}^{2} θ) = {[\sec (\sec^{- 1} 3)]}^{2} + {[cosec ({cosec}^{- 1} 4)]}^{2} - 2$
$= 3^{2} + 4^{2} - 2 = 9 + 16 - 2 = 23$

Thus, the value of $\tan^{2} (\sec^{- 1} 3) + \cot^{2} ({cosec}^{- 1} 4)$ is 23.

The value of tan²(sec⁻¹3) + cot² (cosec⁻¹4) is ____23____.

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Question 19:

If tan⁻¹(cotθ) = 2θ, then θ = __________________.

Answer:

$\tan^{- 1} (\cot θ) = 2 θ \Rightarrow \tan^{- 1} [\tan (\frac{π}{2} - θ)] = 2 θ \Rightarrow \frac{π}{2} - θ = 2 θ \Rightarrow 3 θ = \frac{π}{2} \Rightarrow θ = \frac{π}{6}$

Thus, the value of $θ$ is $\frac{π}{6}$ .

If tan⁻¹(cotθ) = 2θ, then θ = $\frac{π}{6}$ .

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Question 20:

The value of sin^-1 $(\cos \frac{33 π}{5})$ is _________________.

Answer:

$\sin^{- 1} (\cos \frac{33 π}{5}) = \sin^{- 1} [\cos (6 π + \frac{3 π}{5})] = \sin^{- 1} (\cos \frac{3 π}{5})$
$= \sin [\sin (\frac{π}{2} - \frac{3 π}{5})] = \sin [\sin (- \frac{π}{10})] = - \frac{π}{10}$

Thus, the value of $\sin^{- 1} (\cos \frac{33 π}{5})$ is $- \frac{π}{10}$ .

The value of sin⁻¹ $(\cos \frac{33 π}{5})$ is $- \frac{π}{10}$ .

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Question 21:

If tan^-1x + tan^-1y = $\frac{4 π}{5}$ , then cot^-1x + cot^-1y = _________________.

Answer:

We know

$\tan^{- 1} a + \cot^{- 1} a = \frac{π}{2}$ , for all a ∈ R .....(1)

Now,

$\tan^{- 1} x + \tan^{- 1} y = \frac{4 π}{5}$ (Given)

$\Rightarrow \frac{π}{2} - \cot^{- 1} x + \frac{π}{2} - \cot^{- 1} y = \frac{4 π}{5}$ [Using (1)]

$\Rightarrow \cot^{- 1} x + \cot^{- 1} y = π - \frac{4 π}{5}$

$\Rightarrow \cot^{- 1} x + \cot^{- 1} y = \frac{π}{5}$

If tan⁻¹x + tan⁻¹y = $\frac{4 π}{5}$ , then cot⁻¹x + cot⁻¹y = $\frac{π}{5}$ .

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Question 22:

If 3 tan^-1x + cot^-1x = π, then x = ____________________.

Answer:

$3 \tan^{- 1} x + \cot^{- 1} x = π$ (Given)

$\Rightarrow 2 \tan^{- 1} x + \frac{π}{2} = π (\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2})$

$\Rightarrow 2 \tan^{- 1} x = π - \frac{π}{2} = \frac{π}{2}$

$\Rightarrow \tan^{- 1} x = \frac{π}{4}$

$\Rightarrow x = \tan \frac{π}{4} = 1$

Thus, the value of x is 1.

If 3tan⁻¹x + cot⁻¹x = $π$ , then x = ___1___.

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Question 23:

If tan^-12, tan^-13 are measures of two angles of triangle, then the measure of its third angle is _________________.

Answer:

Let the measure of third angle of the triangle be x.

Now,

$\tan^{- 1} 2 + \tan^{- 1} 3 + x = π$ (Angle sum property of triangle)

$\Rightarrow π + \tan^{- 1} (\frac{2 + 3}{1 - 2 \times 3}) + x = π [\tan^{- 1} x + \tan^{- 1} y = π + \tan (\frac{x + y}{1 - x y}), if x y > 1]$

$\Rightarrow \tan^{- 1} (- 1) + x = 0$

$\Rightarrow - \frac{π}{4} + x = 0$

$\Rightarrow x = \frac{π}{4}$

Thus, the measure of third angle of the triangle is $\frac{π}{4}$ .

If tan⁻¹2, tan⁻¹3 are measures of two angles of triangle, then the measure of its third angle is $\frac{π}{4}$ .

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Question 24:

If tan^-1 $\frac{a}{x}$ + tan^-1 $\frac{b}{x} = \frac{π}{2}$ , then x = _________________.

Answer:

We know

$\tan^{- 1} x = \cot^{- 1} \frac{1}{x}$

$∴ \tan^{- 1} \frac{b}{x} = \cot^{- 1} \frac{1}{(\frac{b}{x})} = \cot^{- 1} \frac{x}{b}$ .....(1)
So,

$\tan^{- 1} \frac{a}{x} + \tan^{- 1} \frac{b}{x} = \frac{π}{2}$ (Given)

$\Rightarrow \tan^{- 1} \frac{a}{x} + \cot^{- 1} \frac{x}{b} = \frac{π}{2}$ [Using (1)]

$\Rightarrow \frac{a}{x} = \frac{x}{b} (\tan^{- 1} y + \cot^{- 1} y = \frac{π}{2})$

$\Rightarrow x^{2} = a b$

$\Rightarrow x = \sqrt{a b}$

If tan⁻¹ $\frac{a}{x}$ + tan⁻¹ $\frac{b}{x} = \frac{π}{2}$ , then x = $\sqrt{a b}$ .

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Question 25:

If cos(2sin^-1x) = $\frac{1}{9}$ , then the value of x is ______________.

Answer:

Let $\sin^{- 1} x = θ \Rightarrow \sin θ = x$ .

$∴ \cos (2 \sin^{- 1} x) = \frac{1}{9} \Rightarrow \cos 2 θ = \frac{1}{9} \Rightarrow 1 - 2 \sin^{2} θ = \frac{1}{9}$
$\Rightarrow 2 \sin^{2} θ = 1 - \frac{1}{9} = \frac{8}{9} \Rightarrow \sin^{2} θ = \frac{4}{9} \Rightarrow \sin θ = \pm \frac{2}{3} \Rightarrow x = \pm \frac{2}{3} (\sin θ = x)$

Thus, the value of x is $\pm \frac{2}{3}$ .

If cos(2sin⁻¹x) = $\frac{1}{9}$ , then the value of x is $\pm \frac{2}{3}$ .

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Question 26:

If 0 < x < $\frac{π}{2}$ , then sin^-1 (cos x) + cos^-1(sin x) = ___________________.

Answer:

$\sin^{- 1} (\cos x) + \cos^{- 1} (\sin x)$

$= \sin^{- 1} [\sin (\frac{π}{2} - x)] + \cos^{- 1} [\cos (\frac{π}{2} - x)]$

$= \frac{π}{2} - x + \frac{π}{2} - x$

$= π - 2 x$

If 0 < x < $\frac{π}{2}$ , then sin⁻¹(cos x) + cos⁻¹(sin x) = $π - 2 x$ .

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Question 27:

If $\tan^{- 1} x = \frac{π}{4} - \tan^{- 1} \frac{1}{3},$ then x = ______________________.

Answer:

$\tan^{- 1} x = \frac{π}{4} - \tan^{- 1} \frac{1}{3} \Rightarrow \tan^{- 1} x = \tan^{- 1} 1 - \tan^{- 1} \frac{1}{3} \Rightarrow \tan^{- 1} x = \tan^{- 1} (\frac{1 - \frac{1}{3}}{1 + 1 \times \frac{1}{3}}) [\tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y}), x y > - 1]$
$\Rightarrow \tan^{- 1} x = \tan^{- 1} (\frac{\frac{2}{3}}{\frac{4}{3}}) \Rightarrow \tan^{- 1} x = \tan^{- 1} \frac{1}{2} \Rightarrow x = \frac{1}{2}$

If $\tan^{- 1} x = \frac{π}{4} - \tan^{- 1} \frac{1}{3},$ then x = $\frac{1}{2}$ .

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Question 28:

If tan^-1x + tan^-1 $\frac{1}{2} = \frac{π}{4}$ , then x = _________________.

Answer:

$\tan^{- 1} x + \tan^{- 1} \frac{1}{2} = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{x + \frac{1}{2}}{1 - x \times \frac{1}{2}}) = \frac{π}{4} [\tan^{- 1} a + \tan^{- 1} b = \tan^{- 1} (\frac{a + b}{1 - a b})] \Rightarrow \frac{2 x + 1}{2 - x} = \tan \frac{π}{4} = 1$
$\Rightarrow 2 x + 1 = 2 - x \Rightarrow 3 x = 1 \Rightarrow x = \frac{1}{3}$

Thus, the value of x is $\frac{1}{3}$ .

If tan⁻¹x + tan⁻¹ $\frac{1}{2} = \frac{π}{4}$ , then x = $\frac{1}{3}$ .

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Question 29:

cot $(\frac{π}{4} - 2 \cot^{- 1} 3)$ is equal to ______________________.

Answer:

Let $\cot^{- 1} 3 = θ \Rightarrow \cot θ = 3$ .

$∴ \cot (\frac{π}{4} - 2 \cot^{- 1} 3) = \cot (\frac{π}{4} - 2 θ) = \frac{\cot \frac{π}{4} \cot 2 θ + 1}{\cot 2 θ - \cot \frac{π}{4}} [\cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}]$
$= \frac{\cot 2 θ + 1}{\cot 2 θ - 1} = \frac{\frac{\cot^{2} θ - 1}{2 \cot θ} + 1}{\frac{\cot^{2} θ - 1}{2 \cot θ} - 1} = \frac{\cot^{2} θ - 1 + 2 \cot θ}{\cot^{2} θ - 1 - 2 \cot θ}$
$= \frac{3^{2} - 1 + 2 \times 3}{3^{2} - 1 - 2 \times 3} (\cot θ = 3) = \frac{9 - 1 + 6}{9 - 1 - 6} = \frac{14}{2} = 7$

cot $(\frac{π}{4} - 2 \cot^{- 1} 3)$ is equal to ___7___.

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Question 30:

tan^-1 $(\tan \frac{2 π}{3})$ is equal to __________________.

Answer:

$\tan^{- 1} (\tan \frac{2 π}{3}) = \tan^{- 1} [\tan (π - \frac{π}{3})] = \tan^{- 1} (- \tan \frac{π}{3})$
$= \tan^{- 1} [\tan (- \frac{π}{3})] [\tan (- θ) = - \tan θ] = - \frac{π}{3} [\tan^{- 1} (\tan x) = x, if x \in (- \frac{π}{2}, \frac{π}{2})]$

tan⁻¹ $(\tan \frac{2 π}{3})$ is equal to $- \frac{π}{3}$ .

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Question 31:

If y = 2tan^-1 x+sin^-1 $(\frac{2 x}{1 + x^{2}})$ for all x, then y lies in the interval_________________.

Answer:

We know

$2 \tan^{- 1} x = \{\begin{cases} \sin^{- 1} (\frac{2 x}{1 + x^{2}}), & - 1 \leq x \leq 1 \\ π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}), & x > 1 \\ - π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}), & x < - 1 \end{cases}$

$∴ y = 2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = \{\begin{cases} 4 \tan^{- 1} x, & - 1 \leq x \leq 1 \\ π, & x > 1 \\ - π, & x < - 1 \end{cases}$

For $- 1 \leq x \leq 1$ ,

$- \frac{π}{4} \leq \tan^{- 1} x \leq \frac{π}{4} \Rightarrow - π \leq 4 \tan^{- 1} x \leq π \Rightarrow - π \leq y \leq π . . . . . (1)$

For x > 1, y = $π$ .....(2)

For x < −1, y = $- π$ .....(3)

From (1), (2) and (3), we get

$y \in [- π, π]$ , for all x ∈ R

Thus, the range of y is $[- π, π]$ .

If y = 2tan⁻¹x + sin⁻¹ $(\frac{2 x}{1 + x^{2}})$ for all x, then y lies in the interval $[- π, π]$ .

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Question 32:

The result tan^-1 x-tan^-1 y = tan^-1 $(\frac{x - y}{1 + x y})$ is true when value of xy is __________________.

Answer:

We know

$\tan^{- 1} x - \tan^{- 1} y = \{\begin{cases} \tan^{- 1} (\frac{x - y}{1 + x y}), & if x y > - 1 \\ π + \tan^{- 1} (\frac{x - y}{1 + x y}), & if x > 0, y < 0, x y < - 1 \\ - π + \tan^{- 1} (\frac{x - y}{1 + x y}), & if x < 0, y > 0, x y < - 1 \end{cases}$

Thus, $\tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})$ when the value of xy > −1.

The result tan⁻¹x − tan⁻¹y = tan⁻¹ $(\frac{x - y}{1 + x y})$ is true when value of xy is __greater than − 1___.

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Question 33:

The value of cot^-1(-x) for all x ∊ R in terms of cot^-1 x is _________________.

Answer:

We know

$\cot^{- 1} (- x) = π - \cot^{- 1} x$ , for all x ∈ R

The value of cot⁻¹(−x) for all x ∈ R in terms of cot⁻¹ x is $π - \cot^{- 1} x$ .

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Question 34:

The principal values of $\cos^{- 1} (- \frac{1}{2})$ is ..................

Answer:

We know, $\cos^{- 1} (- θ) = π - \cos^{- 1} (θ)$
$\begin{array}{rcl} ∴ \cos^{- 1} (\frac{- 1}{2}) & = & π - \cos^{- 1} (\frac{1}{2}) \\ = & π - \frac{π}{3} (∵ \cos^{- 1} (\frac{1}{2}) = \frac{π}{3}) \\ = & \frac{2 π}{3} \end{array}$
Thus, the principal value of $\cos^{- 1} (\frac{- 1}{2})$ is $\frac{2 π}{3}$

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Question 35:

The range of the principal value branch of y = sec^–1 x is ........................