Class 12: Inverse Trigonometric Functions – Exercise 4.7 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Evaluate each of the following:

$\displaystyle \text{(i) } \sin^{-1} \Big( \sin \frac{\pi}{6} \Big) \hspace{2.0cm} \text{(ii) } \sin^{-1} \Big( \sin \frac{7\pi}{6} \Big)$

$\displaystyle \text{(iii) } \sin^{-1} \Big( \sin \frac{5\pi}{6} \Big) \hspace{2.0cm} \text{(iv) } \sin^{-1} \Big( \sin \frac{13\pi}{7} \Big)$

$\displaystyle \text{(vi) } \sin^{-1} \Big( \sin\frac{17\pi}{8} \Big) \hspace{2.0cm} \text{(vii) } \sin^{-1} \Big( \sin \frac{-17\pi}{8} \Big)$

$\displaystyle \text{(viii) } \sin^{-1} (\sin 3) \hspace{2.0cm} \text{(ix) } \sin^{-1} (\sin 4)$

$\displaystyle \text{(x) } \sin^{-1} (\sin 12) \hspace{2.0cm} \text{(xi) } \sin^{-1} (\sin 2)$

Answer:

$\displaystyle \textbf{Note: } \ \ \ \ \sin^{-1} ( \sin \theta ) = \theta \text{ for all } \theta \in \Big[ - \frac{\pi}{2} , \frac{\pi}{2} \Big]$

$\displaystyle \text{(i) } \sin^{-1} \Big( \sin \frac{\pi}{6} \Big) = \frac{\pi}{6}$

$\displaystyle \text{(ii) } \sin^{-1} \Big( \sin \frac{7\pi}{6} \Big) = \sin^{-1} \Big( \sin \Big( \pi + \frac{\pi}{6} \Big) \Big) = \sin^{-1} \Big( \sin \Big( -\frac{\pi}{6} \Big) \Big) = -\frac{\pi}{6}$

$\displaystyle \text{(iii) } \sin^{-1} \Big( \sin \frac{5\pi}{6} \Big) = \sin^{-1} \Big( \sin \Big( \pi - \frac{\pi}{6} \Big) \Big) = \sin^{-1} \Big( \sin \Big( \frac{\pi}{6} \Big) \Big) = \frac{\pi}{6}$

$\displaystyle \text{(iv) } \sin^{-1} \Big( \sin \frac{13\pi}{7} \Big) = \sin^{-1} \Big( \sin \Big( 2\pi - \frac{\pi}{7} \Big) \Big) = \sin^{-1} \Big( \sin \Big( -\frac{\pi}{7} \Big) \Big) = -\frac{\pi}{7}$

$\displaystyle \text{(vi) } \sin^{-1} \Big( \sin\frac{17\pi}{8} \Big) = \sin^{-1} \Big( \sin \Big( 2\pi + \frac{\pi}{8} \Big) \Big) = \sin^{-1} \Big( \sin \Big( \frac{\pi}{8} \Big) \Big) = \frac{\pi}{8}$

$\displaystyle \text{(vii)} \sin^{-1} \Big( \sin \frac{-17\pi}{8} \Big) = \sin^{-1} \Big( -\sin \Big( 2\pi + \frac{\pi}{8} \Big) \Big) = \sin^{-1} \Big( -\sin \Big( \frac{\pi}{8} \Big) \Big) = -\frac{\pi}{8}$

$\displaystyle \text{(viii) } \sin^{-1} (\sin 3) = \sin^{-1} (\sin (\pi - 3) )= \pi - 3$

$\displaystyle \text{(ix) } \sin^{-1} (\sin 4) = \sin^{-1} (\sin (\pi - 4) )= \pi - 4$

$\displaystyle \text{(x) } \sin^{-1} (\sin 12) = \sin^{-1} (\sin (-\pi + 12) )= 12 - \pi$

$\displaystyle \text{(xi) } \sin^{-1} (\sin 2) = \sin^{-1} (\sin (\pi - 2) )= \pi - 2$

$\\$

Question 2: Evaluate each of the following:

$\displaystyle \text{(i) } \cos^{-1} \Big( \cos \frac{-\pi}{4} \Big) \hspace{2.0cm} \text{(ii) } \cos^{-1} \Big( \cos \frac{5\pi}{4} \Big)$

$\displaystyle \text{(iii) } \cos^{-1} \Big( \cos \frac{4\pi}{3} \Big) \hspace{2.0cm} \text{(iv) } \cos^{-1} \Big( \cos \frac{13\pi}{6} \Big)$

$\displaystyle \text{(v) } \cos^{-1} (\cos 3) \hspace{2.0cm} \text{(vi) } \cos^{-1} (\cos 4)$

$\displaystyle \text{(vii) } \cos^{-1} (\cos 5) \hspace{2.0cm} \text{(viii) } \cos^{-1} (\cos 12)$

Answer:

$\displaystyle \textbf{Note: } \ \ \ \ \cos^{-1} ( \cos \theta ) = \theta \text{ for all } \theta \in [0, \pi ]$

$\displaystyle \text{(i) } \cos^{-1} \Big( \cos \frac{-\pi}{4} \Big) = \cos^{-1} \Big( \cos \frac{\pi}{4} \Big) = \frac{\pi}{4}$

$\displaystyle \text{(ii) } \cos^{-1} \Big( \cos \frac{5\pi}{4} \Big) = \cos^{-1} \Big( \cos \Big( 2\pi - \frac{3\pi}{4} \Big) \Big) = \cos^{-1} \Big( \cos \frac{3\pi}{4} \Big) = \frac{3\pi}{4}$

$\displaystyle \text{(iii) } \cos^{-1} \Big( \cos \frac{4\pi}{3} \Big) = \cos^{-1} \Big( \cos \Big( 2\pi - \frac{2\pi}{3} \Big) \Big) = \cos^{-1} \Big( \cos \frac{2\pi}{3} \Big) = \frac{2\pi}{3}$

$\displaystyle \text{(iv) } \cos^{-1} \Big( \cos \frac{13\pi}{6} \Big) = \cos^{-1} \Big( \cos \Big( 2\pi + \frac{\pi}{6} \Big) \Big) = \cos^{-1} \Big( \cos \frac{\pi}{6} \Big) = \frac{\pi}{6}$

$\displaystyle \text{(v) } \cos^{-1} (\cos 3) = \cos^{-1} ( \cos ( 2\pi - 3 ) ) = 2\pi - 3$

$\displaystyle \text{(vi) } \cos^{-1} (\cos 4) = \cos^{-1} ( \cos ( 2\pi - 4 ) ) = 2\pi - 4$

$\displaystyle \text{(vii) } \cos^{-1} (\cos 5) = \cos^{-1} ( \cos ( 2\pi - 5 ) ) = 2\pi - 5$

$\displaystyle \text{(viii) } \cos^{-1} (\cos 12) = \cos^{-1} ( \cos ( 4\pi - 12 ) ) = 4\pi - 12$

$\\$

Question 3: Evaluate each of the following:

$\displaystyle \text{(i) } \tan^{-1} \Big( \tan \frac{\pi}{3} \Big) \hspace{2.0cm} \text{(ii) } \tan^{-1} \Big( \tan \frac{6\pi}{7} \Big)$

$\displaystyle \text{(iii) } \tan^{-1} \Big( \tan \frac{7\pi}{6} \Big) \hspace{2.0cm} \text{(iv) } \tan^{-1} \Big( \tan \frac{9\pi}{4} \Big)$

$\displaystyle \text{(v) } \tan^{-1} (\tan 1) \hspace{2.0cm} \text{(v) } \tan^{-1} (\tan 2)$

$\displaystyle \text{(v) } \tan^{-1} (\tan 4) \hspace{2.0cm} \text{(v) } \tan^{-1} (\tan 12)$

Answer:

$\displaystyle \textbf{Note: } \ \ \ \ \tan^{-1} ( \tan \theta ) = \theta \text{ for all } \theta \in \Big( - \frac{\pi}{2} , \frac{\pi}{2} \Big)$

$\displaystyle \text{(i) } \tan^{-1} \Big( \tan \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(ii) } \tan^{-1} \Big( \tan \frac{6\pi}{7} \Big) = \tan^{-1} \Big( \tan \Big( \pi - \frac{\pi}{7} \Big) \Big) = \tan^{-1} \Big( \tan \frac{-\pi}{7} \Big) = - \frac{\pi}{7}$

$\displaystyle \text{(iii) } \tan^{-1} \Big( \tan \frac{7\pi}{6} \Big) = \tan^{-1} \Big( \tan \Big( \pi + \frac{\pi}{6} \Big) \Big) = \tan^{-1} \Big( \tan \frac{\pi}{6} \Big) = \frac{\pi}{6}$

$\displaystyle \text{(iv) } \tan^{-1} \Big( \tan \frac{9\pi}{4} \Big) = \tan^{-1} \Big( \tan \Big( 2\pi + \frac{\pi}{4} \Big) \Big) = \tan^{-1} \Big( \tan \frac{\pi}{4} \Big) = \frac{\pi}{4}$

$\displaystyle \text{(v) } \tan^{-1} (\tan 1) = 1$

$\displaystyle \text{(v) } \tan^{-1} (\tan 2) = \tan^{-1} ( \tan ( -\pi + 2)) = 2 - \pi$

$\displaystyle \text{(v) } \tan^{-1} (\tan 4) = \tan^{-1} ( \tan ( -\pi + 4)) = 4 - \pi$

$\displaystyle \text{(v) } \tan^{-1} (\tan 12) = \tan^{-1} ( \tan ( -4\pi +12)) = 12 - 4\pi$

$\\$

Question 4: Evaluate each of the following:

$\displaystyle \text{(i) } \sec^{-1} \Big( \sec \frac{\pi}{3} \Big) \hspace{2.0cm} \text{(ii) } \sec^{-1} \Big( \sec \frac{2\pi}{3} \Big)$

$\displaystyle \text{(iii) } \sec^{-1} \Big( \sec \frac{5\pi}{4} \Big) \hspace{2.0cm} \text{(iv) } \sec^{-1} \Big( \sec \frac{7\pi}{3} \Big)$

$\displaystyle \text{(v) } \sec^{-1} \Big( \sec \frac{9\pi}{5} \Big) \hspace{2.0cm} \text{(vi) } \sec^{-1} \Big( \sec \frac{-7\pi}{3} \Big)$

$\displaystyle \text{(vii) } \sec^{-1} \Big( \sec \frac{13\pi}{4} \Big) \hspace{2.0cm} \text{v(iii) } \sec^{-1} \Big( \sec \frac{25\pi}{6} \Big)$

Answer:

$\displaystyle \textbf{Note: } \ \ \ \ \sec^{-1} ( \sec \theta ) = \theta \text{ for all } \theta \in \Big[ 0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2}, \pi \Big]$

$\displaystyle \text{(i) } \sec^{-1} \Big( \sec \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(ii) } \sec^{-1} \Big( \sec \frac{2\pi}{3} \Big) = \frac{2\pi}{3}$

$\displaystyle \text{(iii) } \sec^{-1} \Big( \sec \frac{5\pi}{4} \Big) = \sec^{-1} \Big( \sec \Big( 2\pi - \frac{3\pi}{4} \Big) \Big) = \sec^{-1} \Big( \sec \frac{3\pi}{4} \Big) = \frac{3\pi}{4}$

$\displaystyle \text{(iv) } \sec^{-1} \Big( \sec \frac{7\pi}{3} \Big) = \sec^{-1} \Big( \sec \Big( 2\pi + \frac{\pi}{3} \Big) \Big) = \sec^{-1} \Big( \sec \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(v) } \sec^{-1} \Big( \sec \frac{9\pi}{5} \Big) = \sec^{-1} \Big( \sec \Big( 2\pi - \frac{\pi}{5} \Big) \Big) = \sec^{-1} \Big( \sec \frac{\pi}{5} \Big) = \frac{\pi}{5}$

$\displaystyle \text{(vi) } \sec^{-1} \Big( \sec \frac{-7\pi}{3} \Big) = \sec^{-1} \Big( \sec \frac{7\pi}{3} \Big) = \sec^{-1} \Big( \sec \Big( 2\pi + \frac{\pi}{3} \Big) \Big) = \sec^{-1} \Big( \sec \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(vii) } \sec^{-1} \Big( \sec \frac{13\pi}{4} \Big) = \sec^{-1} \Big( \sec \Big( 4\pi - \frac{3\pi}{4} \Big) \Big) = \sec^{-1} \Big( \sec \frac{3\pi}{4} \Big) = \frac{3\pi}{4}$

$\displaystyle \text{v(iii) } \sec^{-1} \Big( \sec \frac{25\pi}{6} \Big) = \sec^{-1} \Big( \sec \Big( 4\pi + \frac{\pi}{6} \Big) \Big) = \sec^{-1} \Big( \sec \frac{\pi}{6} \Big) = \frac{\pi}{6}$

$\\$

Question 5: Evaluate each of the following:

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{\pi}{4} \Big) \hspace{2.0cm} \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{3\pi}{4} \Big)$

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{6\pi}{5} \Big) \hspace{2.0cm} \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{11\pi}{6} \Big)$

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{13\pi}{6} \Big) \hspace{2.0cm} \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{-9\pi}{4} \Big)$

Answer:

$\displaystyle \textbf{Note: } \ \ \ \ \mathrm{cosec}^{-1} ( \mathrm{cosec} \ \theta ) = \theta \text{ for all } \theta \in \Big[ -\frac{\pi}{2}, 0 \Big) \cup \Big( 0, \frac{\pi}{2} \Big]$

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{\pi}{4} \Big) = \frac{\pi}{4}$

$\displaystyle \text{(ii) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{3\pi}{4} \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \Big( \pi - \frac{\pi}{4} \Big) \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{\pi}{4} \Big) = \frac{\pi}{4}$

$\displaystyle \text{(ii) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{6\pi}{5} \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \Big( \pi + \frac{\pi}{5} \Big) \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{-\pi}{5} \Big) = -\frac{\pi}{5}$

$\displaystyle \text{(iv) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{11\pi}{6} \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \Big( 2\pi - \frac{\pi}{6} \Big) \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{-\pi}{6} \Big) = -\frac{\pi}{6}$

$\displaystyle \text{(v) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{13\pi}{6} \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \Big( 2\pi + \frac{\pi}{6} \Big) \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{\pi}{6} \Big) = \frac{\pi}{6}$

$\displaystyle \text{(vi) } \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{-9\pi}{4} \Big) = \mathrm{cosec}^{-1} \Big( -\mathrm{cosec} \Big( 2\pi + \frac{\pi}{4}\Big) \Big) = \mathrm{cosec}^{-1} \Big( -\mathrm{cosec} \frac{\pi}{4} \Big) = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \frac{-\pi}{4} \Big) = -\frac{\pi}{4}$

$\\$

Question 6: Evaluate each of the following:

$\displaystyle \text{(i) } \cot^{-1} \Big( \cot \frac{\pi}{3} \Big) \hspace{2.0cm} \text{(ii) } \cot^{-1} \Big( \cot \frac{4\pi}{3} \Big)$

$\displaystyle \text{(iii) } \cot^{-1} \Big( \cot \frac{9\pi}{4} \Big) \hspace{2.0cm} \text{(iv) } \cot^{-1} \Big( \cot \frac{19\pi}{6} \Big)$

$\displaystyle \text{(v) } \cot^{-1} \Big( \cot \frac{-8\pi}{3} \Big) \hspace{2.0cm} \text{(vi) } \cot^{-1} \Big( \cot \frac{21\pi}{4} \Big)$

Answer:

$\displaystyle \textbf{Note: } \ \ \ \ \cot^{-1} ( \cot \theta ) = \theta \text{ for all } \theta \in (0, \pi )$

$\displaystyle \text{(i) } \cot^{-1} \Big( \cot \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(ii) } \cot^{-1} \Big( \cot \frac{4\pi}{3} \Big) = \cot^{-1} \Big( \cot \Big( \pi + \frac{\pi}{3} \Big) \Big) = \cot^{-1} \Big( \cot \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(iii) } \cot^{-1} \Big( \cot \frac{9\pi}{4} \Big) = \cot^{-1} \Big( \cot \Big( 2\pi + \frac{\pi}{4} \Big) \Big) = \cot^{-1} \Big( \cot \frac{\pi}{4} \Big) = \frac{\pi}{4}$

$\displaystyle \text{(iv) } \cot^{-1} \Big( \cot \frac{19\pi}{6} \Big) = \cot^{-1} \Big( \cot \Big( 3\pi + \frac{\pi}{6} \Big) \Big) = \cot^{-1} \Big( \cot \frac{\pi}{6} \Big) = \frac{\pi}{6}$

$\displaystyle \text{(v) } \cot^{-1} \Big( \cot \frac{-8\pi}{3} \Big) = \cot^{-1} \Big( -\cot \frac{8\pi}{3} \Big) = \cot^{-1} \Big( - \cot \Big( 3\pi - \frac{\pi}{3} \Big) \Big) = \cot^{-1} \Big( \cot \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\displaystyle \text{(vi) } \cot^{-1} \Big( \cot \frac{21\pi}{4} \Big) = \cot^{-1} \Big( \cot \Big( 5\pi + \frac{\pi}{4} \Big) \Big) = \cot^{-1} \Big( \cot \frac{\pi}{4} \Big) = \frac{\pi}{4}$

$\\$

Question 7: Write each of the following in the simplest form:

$\displaystyle \text{(i) } \cot^{-1} \Bigg\{ \frac{a}{\sqrt{x^2-a^2}} \Bigg\} \ , |x| > a$

$\displaystyle \text{(ii) } \tan^{-1} \Big\{ x + \sqrt{1+x^2} \Big\} \ , x \in R$

$\displaystyle \text{(iii) } \tan^{-1} \Big\{ \sqrt{1+x^2} - x \Big\} \ , x \in R$

$\displaystyle \text{(iv) } \tan^{-1} \Bigg\{ \frac{\sqrt{1+x^2} - 1}{x} \Bigg\} \ , x \neq 0$

$\displaystyle \text{(v) } \tan^{-1} \Bigg\{ \frac{\sqrt{1+x^2} + 1}{x} \Bigg\} \ , x \neq 0$

$\displaystyle \text{(vi) } \tan^{-1} \sqrt{ \frac{a - x}{a+x} } \ , -a < x < a$

$\displaystyle \text{(vii) } \tan^{-1} \Bigg\{ \frac{x}{a+ \sqrt{a^2-x^2}} \Bigg\} \ , -a < x < a$

$\displaystyle \text{(viii) } \sin^{-1} \Bigg\{ \frac{x + \sqrt{1-x^2}}{\sqrt{2}} \Bigg\} \ , -\frac{1}{2} < x < \frac{1}{\sqrt{2}}$

$\displaystyle \text{(ix) } \sin^{-1} \Bigg\{ \frac{\sqrt{1+x}+\sqrt{1-x}}{2} \Bigg\} \ , 0 < x< 1$

$\displaystyle \text{(x) } \sin^{-1} \Bigg\{ 2 \tan^{-1} \sqrt{\frac{1-x}{1+x}} \Bigg\} \$

Answer:

$\text{(i) Let } x = a \sec \theta$

$\displaystyle \cot^{-1} \Bigg\{ \frac{a}{\sqrt{x^2-a^2}} \Bigg\} \\ \\ \\ = \cot^{-1} \Bigg\{ \frac{a}{\sqrt{a^2 \sec^2 \theta -a^2}} \Bigg\} \\ \\ \\ = \cot^{-1} \Bigg\{ \frac{a}{a\sqrt{\tan^2 \theta}} \Bigg\} \\ \\ \\ = \cot^{-1} ( \cot \theta ) \\ \\ \\ = \theta \\ \\ \\ = \sec^{-1} \frac{x}{a}$

$\\$

$\text{(ii) Let } x = \cot \theta$

$\displaystyle \tan^{-1} \Big\{ x + \sqrt{1+x^2} \Big\} \\ \\ \\ = \tan^{-1} \Big\{ \cot \theta + \sqrt{1+\cot^2 \theta} \Big\} \\ \\ \\ = \tan^{-1} \{ \cot \theta + \mathrm{cosec} \theta \} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\cos \theta + 1 }{\sin \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{2 \cos^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \cot \frac{\theta}{2} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \tan \Big( \frac{\pi}{2} - \frac{\theta}{2} \Big) \Bigg\} \\ \\ \\ = \Big( \frac{\pi}{2} - \frac{\theta}{2} \Big) \\ \\ \\ = \frac{\pi}{2} - \frac{\cot^{-1} x}{2}$

$\\$

$\text{(iii) Let } x = \cot \theta$

$\displaystyle \tan^{-1} \Big\{ \sqrt{1+x^2} - x \Big\} \\ \\ \\ = \tan^{-1} \Big\{ \sqrt{1+\cot^2 \theta} - \cot \theta \Big\} \\ \\ \\ = \tan^{-1} \{ \mathrm{cosec} \theta - \cot \theta \} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{1- \cos \theta}{\sin \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{2 \sin^2 \frac{\theta}{2} }{2 \sin \frac{\theta}{2}\cos \frac{\theta}{2}} \Bigg\} \\ \\ \\ = \tan^{-1}\Bigg\{ \tan \frac{\theta}{2} \Bigg\} \\ \\ \\ = \frac{\theta}{2} \\ \\ \\ = \frac{\cot^{-1} x}{2}$

$\\$

$\text{(iv) Let } x = \tan \theta$

$\displaystyle \tan^{-1} \Bigg\{ \frac{\sqrt{1+x^2} - 1}{x} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\sqrt{1+\tan^2 \theta} - 1}{\tan \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\sqrt{\sec^2 \theta} - 1}{\tan \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\sec \theta - 1}{\tan \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{1 - \cos \theta }{\sin \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{2 \sin^2 \frac{\theta}{2} }{2 \sin \frac{\theta}{2}\cos \frac{\theta}{2}} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \tan \frac{\theta}{2} \Bigg\} \\ \\ \\ = \frac{\theta}{2} \\ \\ \\ = \frac{\tan^{-1} x}{2}$

$\\$

$\text{(v) Let } x = \tan \theta$

$\displaystyle \tan^{-1} \Bigg\{ \frac{\sqrt{1+x^2} + 1}{x} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\sqrt{1+\tan^2 \theta} + 1}{\tan \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\sqrt{\sec^2 \theta} + 1}{\tan \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\sec \theta + 1}{\tan \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{\cos \theta + 1}{\sin \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{2 \cos^2 \frac{\theta}{2} }{2 \sin \frac{\theta}{2}\cos \frac{\theta}{2}} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \cot \frac{\theta}{2} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \tan \Bigg( \frac{\pi}{2} - \frac{\theta}{2} \Bigg) \Bigg\}\\ \\ \\ = \frac{\pi}{2} - \frac{\tan^{-1} x}{2}$

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$\text{(vi) Let } x = a \cos \theta$

$\displaystyle \tan^{-1} \sqrt{ \frac{a - x}{a+x} } \\ \\ \\ = \tan^{-1} \sqrt{ \frac{a - a \cos \theta}{a+a \cos \theta} } \\ \\ \\ = \tan^{-1} \sqrt{ \frac{1 - \cos \theta}{1+ \cos \theta} } \\ \\ \\ = \tan^{-1} \sqrt{ \frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} } \\ \\ \\ = \tan^{-1} \Bigg\{ \tan \frac{\theta}{2} \Bigg\} \\ \\ \\ = \frac{\theta}{2} \\ \\ \\ = \frac{1}{2} \cos^{-1} \Bigg( \frac{x}{a} \Bigg)$

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$\text{(vii) Let } x = a \sin \theta$

$\displaystyle \tan^{-1} \Bigg\{ \frac{x}{a+ \sqrt{a^2-x^2}} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{a \sin \theta}{a+ \sqrt{a^2-a^2 \sin^2 \theta}} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{a \sin \theta}{a+ a \cos \theta} \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \frac{2 \sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2} } \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{ \tan \frac{\theta}{2} \Bigg\} \\ \\ \\ = \frac{\theta}{2} \\ \\ \\ = \frac{1}{2} \sin^{-1} \Bigg( \frac{x}{a} \Bigg)$

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$\text{(viii) Let } x = \sin \theta$

$\displaystyle \sin^{-1} \Bigg\{ \frac{x + \sqrt{1-x^2}}{\sqrt{2}} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \frac{\sin \theta + \sqrt{1-\sin^2 \theta}}{\sqrt{2}} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \frac{\sin \theta + \cos \theta}{\sqrt{2}} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \cos \frac{\pi}{4}\sin \theta + \sin \frac{\pi}{4} \cos \theta \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \sin \Bigg( \theta + \frac{\pi}{4} \Big) \Bigg\} \\ \\ \\ = \theta + \frac{\pi}{4} \\ \\ \\ = \sin^{-1} x + \frac{\pi}{4}$

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$\text{(ix) Let } x = \cos \theta$

$\displaystyle \sin^{-1} \Bigg\{ \frac{\sqrt{1+x}+\sqrt{1-x}}{2} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{2} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \frac{\sqrt{2 \cos^2 \frac{\theta}{2}}+\sqrt{2 \sin^2 \frac{\theta}{2}}}{2} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \frac{1}{\sqrt{2}} \sin \frac{\theta}{2} + \frac{1}{\sqrt{2}} \cos \frac{\theta}{2} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ \sin \Bigg( \frac{\theta}{2} + \frac{\pi}{4} \Big) \Bigg\} \\ \\ \\ = \frac{\theta}{2} + \frac{\pi}{4} \\ \\ \\ = \frac{\cos^{-1} x}{2} + \frac{\pi}{4}$

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$\text{(x) Let } x = \cos \theta$

$\displaystyle \sin^{-1} \Bigg\{ 2 \tan^{-1} \sqrt{\frac{1-x}{1+x}} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ 2 \tan^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ 2 \tan^{-1} \sqrt{\frac{2 \sin^2 \frac{\theta}{2} }{2 \cos^2 \frac{\theta}{2} } } \Bigg\} \\ \\ \\ = \sin^{-1} \Bigg\{ 2 \tan^{-1} \Bigg( \tan \frac{\theta}{2} \Bigg) \Bigg\} \\ \\ \\ = \sin \theta \\ \\ \\ = \sin ( \cos^{-1} x ) \\ \\ \\ = \sin ( \sin^{-1} ( \sqrt{1-x^2}) ) \\ \\ \\ = \sqrt{1-x^2}$

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Class 12: Inverse Trigonometric Functions – Exercise 4.7

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