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

$\displaystyle \textbf{Question 1:}\ \ \text{Find the principal values of} \sin^{-1}\left(-\frac{1}{2}\right) \hspace{2.0cm}[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }x\in[-1,1],\ \sin^{-1}x\ \text{is an angle }\theta\ \text{in the interval} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \\ \text{whose sine is }x\ \text{i.e. }\sin\theta=x.\ \text{Therefore,}$
$\displaystyle \ \sin^{-1}\left(-\frac{1}{2}\right)=\left\{\text{An angle }\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \text{such that }\sin\theta=-\frac{1}{2}\right\}=-\frac{\pi}{6}$

$\displaystyle \textbf{Question 2:}\ \ \text{Evaluate each of the following:} \ \cos^{-1}\left(\cos\frac{7\pi}{6}\right) \hspace{2.0cm}[\text{CBSE 2009}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ \cos^{-1}\left(\cos\frac{7\pi}{6}\right)\ne\frac{7\pi}{6},\ \text{because }\frac{7\pi}{6}\ \text{does not lie between }0\ \text{and }\pi.$
$\displaystyle \text{Now, }\cos^{-1}\left(\cos\frac{7\pi}{6}\right)=\cos^{-1}\left\{\cos\left(2\pi-\frac{5\pi}{6}\right)\right\}\ \left[\because\ \frac{7\pi}{6}=2\pi-\frac{5\pi}{6}\right]$
$\displaystyle =\cos^{-1}\left(\cos\frac{5\pi}{6}\right)\ \left[\because\ \cos(2\pi-\theta)=\cos\theta\right]$
$\displaystyle =\frac{5\pi}{6}$

$\displaystyle \textbf{Question 3:}\ \ \text{Express each of the following in the simplest form:}$
$\displaystyle \ \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right),\ -\frac{\pi}{2}<x<\frac{\pi}{2} \hspace{5.0cm}[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ \text{We have,}$
$\displaystyle \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan^{-1}\left[\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}\right]$
$\displaystyle =\tan^{-1}\left\{\frac{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}\right\}$
$\displaystyle =\tan^{-1}\left\{\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}\right\}$
$\displaystyle =\tan^{-1}\left\{\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}\right\}$
$\displaystyle =\tan^{-1}\left\{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\frac{\pi}{4}-\frac{x}{2}$
$\displaystyle \left[\because\ -\frac{\pi}{2}<x<\frac{\pi}{2}\Rightarrow-\frac{\pi}{4}<-\frac{x}{2}<\frac{\pi}{4}\Rightarrow0<\frac{\pi}{4}-\frac{x}{2}<\frac{\pi}{2}\right]$

$\displaystyle \textbf{Question 4:}\ \ \text{Prove that:}$
$\displaystyle \ \cot^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}=\frac{x}{2},\ 0<x<\frac{\pi}{2}\qquad [\text{CBSE 2009, 2014, 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ \text{We know that}$
$\displaystyle 1+\sin x=\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}=\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2$
$\displaystyle \therefore\ \cot^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$
$\displaystyle =\cot^{-1}\left\{\frac{\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}+\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}{\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}-\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}\right\}$
$\displaystyle =\cot^{-1}\left\{\frac{\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|+\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|}{\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|-\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|}\right\}\ \left[\because\ \sqrt{z^2}=|z|\right]$
$\displaystyle =\cot^{-1}\left\{\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)+\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)-\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}\right\}\ \left[\because\ 0<\frac{x}{2}<\frac{\pi}{4}:\ \cos\frac{x}{2}>\sin\frac{x}{2}\right]$
$\displaystyle =\cot^{-1}\left\{\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right\}=\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}$

$\displaystyle \textbf{Question 5:}\ \ \text{Prove that:}$
$\displaystyle \ \cot^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}=\frac{\pi}{2}-\frac{x}{2},\ \text{if }\frac{\pi}{2}<x<\pi \hspace{2.0cm}[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ \text{We have,}$
$\displaystyle \cot^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$
$\displaystyle =\cot^{-1}\left\{\frac{\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}+\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}{\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}-\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}\right\}$
$\displaystyle =\cot^{-1}\left\{\frac{\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|+\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|}{\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|-\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|}\right\}$
$\displaystyle =\cot^{-1}\left\{\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)+\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)-\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)}\right\}$
$\displaystyle \left[\because\ \frac{\pi}{2}<x<\pi\Rightarrow\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}\Rightarrow\cos\frac{x}{2}<\sin\frac{x}{2}\right]$
$\displaystyle =\cot^{-1}\left\{\frac{2\sin\frac{x}{2}}{2\cos\frac{x}{2}}\right\}=\cot^{-1}\left(\tan\frac{x}{2}\right)$
$\displaystyle =\cot^{-1}\left\{\cot\left(\frac{\pi}{2}-\frac{x}{2}\right)\right\}=\frac{\pi}{2}-\frac{x}{2}$
$\displaystyle \left[\because\ \frac{\pi}{2}<x<\pi\Rightarrow0<\frac{\pi}{2}-\frac{x}{2}<\frac{\pi}{4}\right]$

$\displaystyle \textbf{Question 6:}\ \ \text{Prove that:}$
$\displaystyle \ \tan^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x,\ 0<x<1 \hspace{2.0cm}$
$\displaystyle \text{Answer:}$
$\displaystyle \ \text{Putting }x=\cos2\theta,\ \text{we obtain}$
$\displaystyle \tan^{-1}\left\{\frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}\right\}$
$\displaystyle =\tan^{-1}\left\{\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\right\}$
$\displaystyle =\tan^{-1}\left\{\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\right\}$
$\displaystyle \left[\because\ 0<x<1\Rightarrow0<\cos2\theta<1\Rightarrow0<2\theta<\frac{\pi}{2}\Rightarrow0<\theta<\frac{\pi}{4}\right]$
$\displaystyle =\tan^{-1}\left\{\frac{1-\tan\theta}{1+\tan\theta}\right\}=\tan^{-1}\left\{\tan\left(\frac{\pi}{4}-\theta\right)\right\}$
$\displaystyle =\frac{\pi}{4}-\theta$
$\displaystyle =\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x$
$\displaystyle \left[\because\ \cos2\theta=x\Rightarrow2\theta=\cos^{-1}x\Rightarrow\theta=\frac{1}{2}\cos^{-1}x\right]$

$\displaystyle \textbf{Question 7:}\ \ \text{If }\sin\left\{\cot^{-1}(x+1)\right\}=\cos\left(\tan^{-1}x\right),\ \text{then find }x.\qquad [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \sin\left\{\cot^{-1}(x+1)\right\}=\cos\left(\tan^{-1}x\right)$
$\displaystyle \Rightarrow\ \sin\left\{\sin^{-1}\left(\frac{1}{\sqrt{x^2+2x+2}}\right)\right\}=\cos\left\{\cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right\}$
$\displaystyle \Rightarrow\ \frac{1}{\sqrt{x^2+2x+2}}=\frac{1}{\sqrt{1+x^2}}$
$\displaystyle \Rightarrow\ \sqrt{x^2+2x+2}=\sqrt{1+x^2}\ \Rightarrow\ x^2+2x+2=x^2+1$
$\displaystyle \Rightarrow\ 2x=-1\ \Rightarrow\ x=-\frac{1}{2}$
$\displaystyle \text{Hence, }x=-\frac{1}{2}\ \text{is a root of the given equation.}$

$\displaystyle \textbf{Question 8:}\ \ \text{Solve the following equation for }x:$
$\displaystyle \ \cos\left(\tan^{-1}x\right)=\sin\left(\cot^{-1}\frac{3}{4}\right)\qquad [\text{CBSE 2013, 2014, 2017}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ \text{We have,}$
$\displaystyle \cos\left(\tan^{-1}x\right)=\sin\left(\cot^{-1}\frac{3}{4}\right)$
$\displaystyle \Rightarrow\ \cos\left\{\cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right\}=\sin\left\{\sin^{-1}\left(\frac{4}{5}\right)\right\}$
$\displaystyle \Rightarrow\ \frac{1}{\sqrt{1+x^2}}=\frac{4}{5}$
$\displaystyle \Rightarrow\ 5=4\sqrt{1+x^2}\ \Rightarrow\ 25=16(1+x^2)$
$\displaystyle \Rightarrow\ 25=16+16x^2\ \Rightarrow\ 16x^2=9\ \Rightarrow\ x=\pm\frac{3}{4}$

$\displaystyle \textbf{Question 9:}\ \ \text{Prove that:}$
$\displaystyle \tan\left\{\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\left(\frac{a}{b}\right)\right\}+\tan\left\{\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\left(\frac{a}{b}\right)\right\}=\frac{2b}{a}\qquad [\text{CBSE 2017}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\cos^{-1}\left(\frac{a}{b}\right)=\theta.\ \text{Then, }\cos\theta=\frac{a}{b}$
$\displaystyle \therefore\ \text{LHS}=\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)+\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$\displaystyle \Rightarrow\ \text{LHS}=\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}+\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}$
$\displaystyle \Rightarrow\ \text{LHS}=\frac{\left(1+\tan\frac{\theta}{2}\right)^2+\left(1-\tan\frac{\theta}{2}\right)^2}{1-\tan^2\frac{\theta}{2}}$
$\displaystyle =2\left(\frac{1+\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\right)=\frac{2}{\cos\theta}=\frac{2b}{a}=\text{RHS.}$

$\displaystyle \textbf{Question 10:}\ \ \text{Find the value of }\cot\left(\tan^{-1}a+\cot^{-1}a\right).\qquad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that }\tan^{-1}a+\cot^{-1}a=\frac{\pi}{2}.$
$\displaystyle \therefore\ \cot\left(\tan^{-1}a+\cot^{-1}a\right)=\cot\frac{\pi}{2}=0$

$\displaystyle \textbf{Question 11:}\ \ \text{If }\sin\left(\sin^{-1}\frac{1}{5}+\cos^{-1}x\right)=1,\ \text{then find the value of }x.\qquad [\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \sin\left(\sin^{-1}\frac{1}{5}+\cos^{-1}x\right)=1$
$\displaystyle \Rightarrow\ \sin^{-1}\frac{1}{5}+\cos^{-1}x=\sin^{-1}1$
$\displaystyle \Rightarrow\ \sin^{-1}\frac{1}{5}+\cos^{-1}x=\frac{\pi}{2}$
$\displaystyle \Rightarrow\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}\frac{1}{5}$
$\displaystyle \Rightarrow\ \cos^{-1}x=\cos^{-1}\frac{1}{5}\ \left[\because\ \sin^{-1}\frac{1}{5}+\cos^{-1}\frac{1}{5}=\frac{\pi}{2}\right]$
$\displaystyle \Rightarrow\ x=\frac{1}{5}$

$\displaystyle \textbf{Question 12:}\ \ \text{If }\left(\tan^{-1}x\right)^2+\left(\cot^{-1}x\right)^2=\frac{5\pi^2}{8},\ \text{then find }x.\qquad [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \left(\tan^{-1}x\right)^2+\left(\cot^{-1}x\right)^2=\frac{5\pi^2}{8}$
$\displaystyle \Rightarrow\ \left(\tan^{-1}x\right)^2+\left(\cot^{-1}x\right)^2+2\tan^{-1}x\cot^{-1}x-2\tan^{-1}x\cot^{-1}x=\frac{5\pi^2}{8}$
$\displaystyle \Rightarrow\ \left(\tan^{-1}x+\cot^{-1}x\right)^2-2\tan^{-1}x\cot^{-1}x=\frac{5\pi^2}{8}$
$\displaystyle \Rightarrow\ \frac{\pi^2}{4}-2\tan^{-1}x\left(\frac{\pi}{2}-\tan^{-1}x\right)=\frac{5\pi^2}{8}\ \left[\because\ \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\right]$
$\displaystyle \Rightarrow\ \frac{\pi^2}{4}-\pi\tan^{-1}x+2\left(\tan^{-1}x\right)^2=\frac{5\pi^2}{8}$
$\displaystyle \Rightarrow\ 2\left(\tan^{-1}x\right)^2-\pi\tan^{-1}x-\frac{3\pi^2}{8}=0$
$\displaystyle \Rightarrow\ 16\left(\tan^{-1}x\right)^2-8\pi\tan^{-1}x-3\pi^2=0$
$\displaystyle \Rightarrow\ 16\left(\tan^{-1}x\right)^2-12\pi\tan^{-1}x+4\pi\tan^{-1}x-3\pi^2=0$
$\displaystyle \Rightarrow\ 4\tan^{-1}x\left(4\tan^{-1}x-3\pi\right)+\pi\left(4\tan^{-1}x-3\pi\right)=0$
$\displaystyle \Rightarrow\ \left(4\tan^{-1}x-3\pi\right)\left(4\tan^{-1}x+\pi\right)=0$
$\displaystyle \Rightarrow\ 16\left(\tan^{-1}x-\frac{3\pi}{4}\right)\left(\tan^{-1}x+\frac{\pi}{4}\right)=0$
$\displaystyle \Rightarrow\ \tan^{-1}x+\frac{\pi}{4}=0\ \left[\because\ -\frac{\pi}{2}<\tan^{-1}x<\frac{\pi}{2},\ \tan^{-1}x-\frac{3\pi}{4}\ne0\right]$
$\displaystyle \Rightarrow\ \tan^{-1}x=-\frac{\pi}{4}\ \Rightarrow\ x=\tan\left(-\frac{\pi}{4}\right)=-1$

$\displaystyle \textbf{Question 13:}\ \ \text{Prove that: }\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi\qquad [\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\tan^{-1}1+\left(\tan^{-1}2+\tan^{-1}3\right)$
$\displaystyle =\frac{\pi}{4}+\frac{3\pi}{4}\qquad [\text{See Example 2}]$
$\displaystyle =\pi$

$\displaystyle \textbf{Question 14:}\ \ \text{Prove that:}$
$\displaystyle (i)\ \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}=\frac{\pi}{4}\qquad [\text{CBSE 2011, 2013}]$
$\displaystyle (ii)\ \tan^{-1}\frac{3}{4}+\tan^{-1}\frac{3}{5}-\tan^{-1}\frac{8}{19}=\frac{\pi}{4}\qquad [\text{CBSE 2012}]$
$\displaystyle (iii)\ \tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{8}=\frac{\pi}{4}\qquad [\text{CBSE 2008, 2010, 2016}]$
$\displaystyle (iv)\ \cot^{-1}7+\cot^{-1}8+\cot^{-1}18=\cot^{-1}3\qquad [\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{LHS}=\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}$
$\displaystyle =\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2}\cdot\frac{1}{5}}\right)+\tan^{-1}\frac{1}{8}$
$\displaystyle =\tan^{-1}\frac{7}{9}+\tan^{-1}\frac{1}{8}=\tan^{-1}\left(\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{9}\cdot\frac{1}{8}}\right)$
$\displaystyle =\tan^{-1}\frac{65}{65}=\tan^{-1}1=\frac{\pi}{4}=\text{RHS.}$
$\displaystyle (ii)\ \text{LHS}=\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{3}{5}-\tan^{-1}\frac{8}{19}$
$\displaystyle =\tan^{-1}\frac{27}{11}-\tan^{-1}\frac{8}{19}=\tan^{-1}\left(\frac{\frac{27}{11}-\frac{8}{19}}{1+\frac{27}{11}\cdot\frac{8}{19}}\right)$
$\displaystyle =\tan^{-1}\frac{425}{425}=\tan^{-1}1=\frac{\pi}{4}=\text{RHS.}$
$\displaystyle (iii)\ \text{LHS}=\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{8}$
$\displaystyle =\tan^{-1}\frac{6}{17}+\tan^{-1}\frac{11}{23}=\tan^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\cdot\frac{11}{23}}\right)$
$\displaystyle =\tan^{-1}\frac{325}{325}=\tan^{-1}1=\frac{\pi}{4}=\text{RHS.}$
$\displaystyle (iv)\ \text{LHS}=\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$
$\displaystyle =\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{18}\ \left[\because\ \cot^{-1}x=\tan^{-1}\frac{1}{x},\ x>0\right]$
$\displaystyle =\tan^{-1}\frac{3}{11}+\tan^{-1}\frac{1}{18}=\tan^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\cdot\frac{1}{18}}\right)$
$\displaystyle =\tan^{-1}\frac{65}{195}=\tan^{-1}\frac{1}{3}=\cot^{-1}3=\text{RHS.}$

$\displaystyle \textbf{Question 15:}\ \ \text{Solve the following equations:}$
$\displaystyle (i)\ \tan^{-1}\left(\frac{x-1}{x-2}\right)+\tan^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\qquad [\text{CBSE 2010}]$
$\displaystyle (ii)\ \tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}\qquad [\text{CBSE 2009, 2012}]$

$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{We have,}$
$\displaystyle \tan^{-1}\left(\frac{x-1}{x-2}\right)+\tan^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{x-1}{x-2}\right)+\tan^{-1}\left(\frac{x+1}{x+2}\right)=\tan^{-1}1$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{x-1}{x-2}\right)=\tan^{-1}1-\tan^{-1}\left(\frac{x+1}{x+2}\right)$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{x-1}{x-2}\right)=\tan^{-1}\left(\frac{1-\frac{x+1}{x+2}}{1+\frac{x+1}{x+2}}\right)$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{x-1}{x-2}\right)=\tan^{-1}\left(\frac{x+2-x-1}{x+2+x+1}\right)$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{x-1}{x-2}\right)=\tan^{-1}\left(\frac{1}{2x+3}\right)$
$\displaystyle \Rightarrow\ \frac{x-1}{x-2}=\frac{1}{2x+3}$
$\displaystyle \Rightarrow\ (2x+3)(x-1)=x-2\Rightarrow2x^2+x-3=x-2\Rightarrow2x^2-1=0$
$\displaystyle \Rightarrow\ x=\pm\frac{1}{\sqrt{2}}$
$\displaystyle (ii)\ \text{We have,}$
$\displaystyle \tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{2x+3x}{1-2x\cdot3x}\right)=\tan^{-1}1,\ \text{if }6x^2<1$
$\displaystyle \Rightarrow\ \frac{5x}{1-6x^2}=1,\ \text{if }6x^2<1$
$\displaystyle \Rightarrow\ 6x^2+5x-1=0\ \text{and}\ x^2<\frac{1}{6}$
$\displaystyle \Rightarrow\ (6x-1)(x+1)=0\ \text{and}\ -\frac{1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}}$
$\displaystyle \Rightarrow\ x=-1,\ \frac{1}{6}\ \text{and}\ -\frac{1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}}$
$\displaystyle \Rightarrow\ x=\frac{1}{6}$

$\displaystyle \textbf{Question 16:}\ \ \text{Prove that: }\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}=\tan^{-1}\left(\frac{77}{36}\right)\qquad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}$
$\displaystyle =\sin^{-1}\left\{\frac{8}{17}\sqrt{1-\left(\frac{3}{5}\right)^2}+\frac{3}{5}\sqrt{1-\left(\frac{8}{17}\right)^2}\right\}$
$\displaystyle =\sin^{-1}\left\{\frac{8}{17}\times\frac{4}{5}+\frac{3}{5}\times\frac{15}{17}\right\}=\sin^{-1}\left(\frac{77}{85}\right)=\tan^{-1}\left(\frac{77}{36}\right)$

$\displaystyle \textbf{Question 17:}\ \ \text{Prove that: }\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{56}{65}\qquad [\text{CBSE 2010, 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}$
$\displaystyle =\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{3}{5}\ \left[\because\ \cos^{-1}\frac{12}{13}=\sin^{-1}\frac{5}{13}\right]$
$\displaystyle =\sin^{-1}\left\{\frac{5}{13}\sqrt{1-\left(\frac{3}{5}\right)^2}+\frac{3}{5}\sqrt{1-\left(\frac{5}{13}\right)^2}\right\}$
$\displaystyle =\sin^{-1}\left\{\frac{5}{13}\times\frac{4}{5}+\frac{3}{5}\times\frac{12}{13}\right\}=\sin^{-1}\left(\frac{56}{65}\right)$

$\displaystyle \textbf{Question 18:}\ \ \text{Prove that:}$
$\displaystyle (i)\ \sin^{-1}\frac{4}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{16}{65}=\frac{\pi}{2}\qquad [\text{CBSE 2009}]$
$\displaystyle (ii)\ \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{8}{17}=\cos^{-1}\frac{36}{85}\qquad [\text{CBSE 2010}]$
$\displaystyle (iii)\ \sin^{-1}\frac{3}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}=\sin^{-1}\frac{56}{65}\qquad [\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Using }\sin^{-1}x+\sin^{-1}y=\sin^{-1}\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\},\ \text{we obtain}$
$\displaystyle (i)\ \sin^{-1}\frac{4}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{16}{65}$
$\displaystyle =\sin^{-1}\frac{63}{65}+\sin^{-1}\frac{16}{65}$
$\displaystyle =\cos^{-1}\frac{16}{65}+\sin^{-1}\frac{16}{65}\ \left[\because\ \sin^{-1}\frac{63}{65}=\cos^{-1}\frac{16}{65}\right]$
$\displaystyle =\frac{\pi}{2}$
$\displaystyle (ii)\ \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{8}{17}=\sin^{-1}\frac{77}{85}$
$\displaystyle =\cos^{-1}\sqrt{1-\left(\frac{77}{85}\right)^2}=\cos^{-1}\frac{36}{85}$
$\displaystyle (iii)\ \sin^{-1}\frac{3}{5}+\cos^{-1}\frac{12}{13}$
$\displaystyle =\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\ \left[\because\ \cos^{-1}\frac{12}{13}=\sin^{-1}\frac{5}{13}\right]$
$\displaystyle =\sin^{-1}\left\{\frac{3}{5}\sqrt{1-\left(\frac{5}{13}\right)^2}+\frac{5}{13}\sqrt{1-\left(\frac{3}{5}\right)^2}\right\}$
$\displaystyle =\sin^{-1}\left\{\frac{3}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{4}{5}\right\}=\sin^{-1}\frac{56}{65}$
$\displaystyle =\cos^{-1}\sqrt{1-\left(\frac{56}{65}\right)^2}=\cos^{-1}\frac{33}{65}$

$\displaystyle \textbf{Question 19:}\ \ \text{Prove that: }\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}\qquad [\text{CBSE 2010, 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}$
$\displaystyle =\cos^{-1}\left\{\frac{4}{5}\cdot\frac{12}{13}-\sqrt{1-\left(\frac{4}{5}\right)^2}\sqrt{1-\left(\frac{12}{13}\right)^2}\right\}$
$\displaystyle =\cos^{-1}\left\{\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}\right\}$
$\displaystyle =\cos^{-1}\left(\frac{48}{65}-\frac{15}{65}\right)=\cos^{-1}\frac{33}{65}$

$\displaystyle \textbf{Question 20:}\ \ \text{If }\cos^{-1}\frac{x}{a}+\cos^{-1}\frac{y}{b}=\alpha, \text{prove that:} \\ \frac{x^2}{a^2}-\frac{2xy}{ab}\cos\alpha+\frac{y^2}{b^2}=\sin^2\alpha.\qquad [\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \cos^{-1}\frac{x}{a}+\cos^{-1}\frac{y}{b}=\alpha$
$\displaystyle \Rightarrow\ \cos^{-1}\left\{\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}\right\}=\alpha$
$\displaystyle \Rightarrow\ \frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}=\cos\alpha$
$\displaystyle \Rightarrow\ \frac{xy}{ab}-\cos\alpha=\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}$
$\displaystyle \Rightarrow\ \left(\frac{xy}{ab}-\cos\alpha\right)^2=\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)$
$\displaystyle \Rightarrow\ \frac{x^2y^2}{a^2b^2}-\frac{2xy}{ab}\cos\alpha+\cos^2\alpha=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2y^2}{a^2b^2}$
$\displaystyle \Rightarrow\ \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos\alpha=1-\cos^2\alpha$
$\displaystyle \Rightarrow\ \frac{x^2}{a^2}-\frac{2xy}{ab}\cos\alpha+\frac{y^2}{b^2}=\sin^2\alpha$

$\displaystyle \textbf{Question 21:}\ \ \text{Prove that:}$
$\displaystyle \tan^{-1}x+\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),\ |x|<\frac{1}{\sqrt{3}}\qquad [\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \tan^{-1}x+\tan^{-1}\left(\frac{2x}{1-x^2}\right)$
$\displaystyle =\tan^{-1}\left\{\frac{x+\frac{2x}{1-x^2}}{1-x\cdot\frac{2x}{1-x^2}}\right\}\ \left[\because\ \tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right),\ xy<1\right]$
$\displaystyle =\tan^{-1}\left\{\frac{\frac{x(1-x^2)+2x}{1-x^2}}{\frac{1-x^2-2x^2}{1-x^2}}\right\}$
$\displaystyle =\tan^{-1}\left(\frac{x-x^3+2x}{1-x^2-2x^2}\right)=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$

$\displaystyle \textbf{Question 22:}\ \ \text{Evaluate: }\tan\left(2\tan^{-1}\frac{1}{5}\right)\qquad [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \tan\left(2\tan^{-1}\frac{1}{5}\right)=\tan\left\{\tan^{-1}\left(\frac{2\times\frac{1}{5}}{1-\left(\frac{1}{5}\right)^2}\right)\right\}$
$\displaystyle =\tan\left(\tan^{-1}\frac{5}{12}\right)=\frac{5}{12}$

$\displaystyle \textbf{Question 23:}\ \ \text{Prove that:}$
$\displaystyle \tan\frac{1}{2}\left\{\sin^{-1}\left(\frac{2x}{1+x^2}\right)+\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\frac{x+y}{1-xy}, \\ \text{if }|x|<1,y>0\ \text{and }xy<1. \hspace{5.0cm} [\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that }\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)=2\tan^{-1}y\ \text{for all }y\ge0$
$\displaystyle \text{and, }\sin^{-1}\left(\frac{2x}{1+x^2}\right)=2\tan^{-1}x\ \text{for all }x\in[-1,1]$
$\displaystyle \therefore\ \text{LHS}=\tan\frac{1}{2}\left\{\sin^{-1}\left(\frac{2x}{1+x^2}\right)+\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}$
$\displaystyle \Rightarrow\ \text{LHS}=\tan\frac{1}{2}\left\{2\tan^{-1}x+2\tan^{-1}y\right\}$
$\displaystyle \Rightarrow\ \text{LHS}=\tan\left(\tan^{-1}x+\tan^{-1}y\right)$
$\displaystyle \Rightarrow\ \text{LHS}=\tan\left\{\tan^{-1}\left(\frac{x+y}{1-xy}\right)\right\}\qquad [\because\ xy<1]$
$\displaystyle \Rightarrow\ \text{LHS}=\frac{x+y}{1-xy}=\text{RHS}$

$\displaystyle \textbf{Question 24:}\ \ \text{Prove that: }\tan^{-1}\sqrt{x}=\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right),\ x\in[0,1]\qquad [\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\cos^{-1}\left(\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}\right)$
$\displaystyle =\frac{1}{2}\times2\tan^{-1}\sqrt{x}=\tan^{-1}\sqrt{x}$
$\displaystyle \text{Putting }x=\tan^2\theta,\ \text{we obtain}$
$\displaystyle \text{RHS}=\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$
$\displaystyle =\frac{1}{2}\cos^{-1}(\cos2\theta)=\theta=\tan^{-1}\sqrt{x}=\text{LHS}$

$\displaystyle \textbf{Question 25:}\ \ \text{Prove the following:}$
$\displaystyle \ 2\tan^{-1}\frac{1}{5}+\sec^{-1}\frac{5\sqrt{2}}{7}+2\tan^{-1}\frac{1}{8}=\frac{\pi}{4}\qquad [\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ 2\tan^{-1}\frac{1}{5}+\sec^{-1}\frac{5\sqrt{2}}{7}+2\tan^{-1}\frac{1}{8}$
$\displaystyle =2\left\{\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}\right\}+\sec^{-1}\frac{5\sqrt{2}}{7}$
$\displaystyle =2\tan^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\cdot\frac{1}{8}}\right)+\tan^{-1}\sqrt{\left(\frac{5\sqrt{2}}{7}\right)^2-1}\ \left[\because\ \sec^{-1}x=\tan^{-1}\sqrt{x^2-1}\right]$
$\displaystyle =2\tan^{-1}\frac{13}{39}+\tan^{-1}\frac{1}{7}$
$\displaystyle =2\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}$
$\displaystyle =\tan^{-1}\left(\frac{2\times\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}\right)+\tan^{-1}\frac{1}{7}$
$\displaystyle =\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{1}{7}$
$\displaystyle =\tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\cdot\frac{1}{7}}\right)=\tan^{-1}1=\frac{\pi}{4}$

$\displaystyle \textbf{Question 26:}\ \ \text{Solve for }x:\ 2\tan^{-1}(\cos x)=\tan^{-1}(2\mathrm{cosec}\ x). [\text{CBSE 2009, 2010 C, 2014, 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle 2\tan^{-1}(\cos x)=\tan^{-1}(2\mathrm{cosec}\ x)$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{2\cos x}{1-\cos^2x}\right)=\tan^{-1}(2\mathrm{cosec}\ x)$
$\displaystyle \Rightarrow\ \frac{2\cos x}{\sin^2x}=2\mathrm{cosec}\ x\Rightarrow\cos x=\sin x\Rightarrow\tan x=1$
$\displaystyle \Rightarrow\ \tan x=\tan\frac{\pi}{4}\Rightarrow x=n\pi+\frac{\pi}{4},\ n\in Z$

$\displaystyle \textbf{Question 27:}\ \ \text{Solve the following equations:}$
$\displaystyle \ \sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x\qquad [\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \ \text{We have,}$
$\displaystyle \sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$
$\displaystyle \Rightarrow\ \sin^{-1}x+\sin^{-1}(1-x)=\frac{\pi}{2}-\sin^{-1}x\ \ \left[\because\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\right]$
$\displaystyle \Rightarrow\ \sin^{-1}(1-x)=\frac{\pi}{2}-2\sin^{-1}x$
$\displaystyle \Rightarrow\ \sin\left\{\sin^{-1}(1-x)\right\}=\sin\left(\frac{\pi}{2}-2\sin^{-1}x\right)$
$\displaystyle \Rightarrow\ 1-x=\cos\left(2\sin^{-1}x\right)$
$\displaystyle \Rightarrow\ 1-x=\cos\left\{\cos^{-1}(1-2x^2)\right\}\ \ \left[\because\ 2\sin^{-1}x=\cos^{-1}(1-2x^2)\right]$
$\displaystyle \Rightarrow\ 1-x=1-2x^2\Rightarrow2x^2-x=0\Rightarrow x(2x-1)=0\Rightarrow x=0\ \text{or,}\ x=\frac{1}{2}$
$\displaystyle \text{Clearly, these values satisfy the given equation.}$
$\displaystyle \text{Hence, }x=0,\frac{1}{2}\ \text{are the roots of the given equation.}$

$\displaystyle \textbf{Question 28:}\ \ \text{Solve for }x: \\ \tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi}{3},\ -1<x<1.\qquad [\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that}$
$\displaystyle \tan^{-1}\left(\frac{1}{x}\right)=\begin{cases}\cot^{-1}x,&\text{if }x>0\\-\pi+\cot^{-1}x,&\text{if }x<0\end{cases}$
$\displaystyle \text{i.e. }\cot^{-1}x=\begin{cases}\tan^{-1}\left(\frac{1}{x}\right),&\text{if }x>0\\\pi+\tan^{-1}\left(\frac{1}{x}\right),&\text{if }x<0\end{cases}$
$\displaystyle \text{CASE I: When }0<x<1:$
$\displaystyle \cot^{-1}\left(\frac{1-x^2}{2x}\right)=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$
$\displaystyle \text{Given equation is}$
$\displaystyle \tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi}{3}$
$\displaystyle \Rightarrow\ 2\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}$
$\displaystyle \Rightarrow\ 4\tan^{-1}x=\frac{\pi}{3}\Rightarrow\tan^{-1}x=\frac{\pi}{12}\Rightarrow x=\tan\frac{\pi}{12}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\displaystyle \text{CASE II: When }-1<x<0:$
$\displaystyle \cot^{-1}\left(\frac{1-x^2}{2x}\right)=\pi+\tan^{-1}\left(\frac{2x}{1-x^2}\right)$
$\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{2x}{1-x^2}\right)+\pi+\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}$
$\displaystyle \Rightarrow\ 2\tan^{-1}\left(\frac{2x}{1-x^2}\right)=-\frac{2\pi}{3}$
$\displaystyle \Rightarrow\ 4\tan^{-1}x=-\frac{2\pi}{3}\Rightarrow\tan^{-1}x=-\frac{\pi}{6}\Rightarrow x=\tan\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}$
$\displaystyle \text{CASE III: When }x=0:$
$\displaystyle \text{LHS}=\tan^{-1}(0)+\cot^{-1}(\infty)=\frac{\pi}{2},\ \text{and RHS}=\frac{\pi}{3}$
$\displaystyle \therefore\ x=0\ \text{is not a solution.}$
$\displaystyle \text{Hence, }x=\frac{\sqrt{3}-1}{\sqrt{3}+1}\ \text{and}\ x=-\frac{1}{\sqrt{3}}\ \text{are the solutions.}$