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

$\displaystyle \textbf{Question 1. } \text{The value of }\mathrm{cosec}\left(\sin^{-1}\left(\frac{-1}{2}\right)\right)-\sec\left(\cos^{-1}\left(\frac{-1}{2}\right)\right)\text{ is} \ \$
$\displaystyle \text{equal to \hspace{9.2cm} [ISC 2024]} \ \$
$\displaystyle \text{(a) }-4 \ \$
$\displaystyle \text{(b) }0 \ \$
$\displaystyle \text{(c) }-1 \ \$
$\displaystyle \text{(d) }4 \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) We have, } \mathrm{cosec}\left[\sin^{-1}\left(\frac{-1}{2}\right)-\sec\left(\cos^{-1}\left(\frac{-1}{2}\right)\right)\right]$
$\displaystyle = \mathrm{cosec}\left[\sin^{-1}\left\{\sin\left(-\frac{\pi}{6}\right)\right\}\right]-\sec\left[\cos^{-1}\left\{\cos\left(\frac{2\pi}{3}\right)\right\}\right]$
$\displaystyle \left[\because -\sin\frac{\pi}{6}=\sin\left(-\frac{\pi}{6}\right)=\frac{-1}{2}\ \text{and}\ \cos\left(\frac{2\pi}{3}\right)=\frac{-1}{2}\right]$
$\displaystyle = \mathrm{cosec}\left[-\frac{\pi}{6}\right]-\sec\left[\frac{2\pi}{3}\right]$
$\displaystyle = -2-(-2)=0$
$\\$

$\displaystyle \textbf{Question 2. }\text{Solve for }x:\ \sin^{-1}\left(\frac{x}{2}\right)+\cos^{-1}x=\frac{\pi}{6}.\text{ \hspace{3.2cm} [ISC 2024]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \sin^{-1}\frac{x}{2} + \cos^{-1}x = \frac{\pi}{6}$
$\displaystyle \Rightarrow \cos^{-1}x = \frac{\pi}{6} - \sin^{-1}\frac{x}{2}$
$\displaystyle \Rightarrow x = \cos\left(\frac{\pi}{6} - \sin^{-1}\frac{x}{2}\right)$
$\displaystyle \Rightarrow x = \cos\frac{\pi}{6}\cos\left(\sin^{-1}\frac{x}{2}\right) + \sin\frac{\pi}{6}\sin\left(\sin^{-1}\frac{x}{2}\right)$
$\displaystyle \Rightarrow x = \frac{\sqrt{3}}{2}\cos\left(\sin^{-1}\frac{x}{2}\right) + \frac{1}{2}\cdot\frac{x}{2}$
$\displaystyle \text{Let } \sin^{-1}\frac{x}{2} = \theta \Rightarrow \frac{x}{2} = \sin\theta$
$\displaystyle \Rightarrow \cos\theta = \sqrt{1 - \frac{x^{2}}{4}}$
$\displaystyle \Rightarrow \theta = \cos^{-1}\sqrt{1 - \frac{x^{2}}{4}}$
$\displaystyle \therefore x = \frac{\sqrt{3}}{2}\cos\left(\cos^{-1}\sqrt{1 - \frac{x^{2}}{4}}\right) + \frac{x}{4}$
$\displaystyle \Rightarrow x = \frac{\sqrt{3}}{2}\sqrt{1 - \frac{x^{2}}{4}} + \frac{x}{4}$
$\displaystyle \Rightarrow x - \frac{x}{4} = \frac{\sqrt{3}}{2}\sqrt{1 - \frac{x^{2}}{4}}$
$\displaystyle \Rightarrow \frac{3x}{4} = \frac{\sqrt{3}}{2}\sqrt{1 - \frac{x^{2}}{4}}$
$\displaystyle \Rightarrow \frac{9x^{2}}{16} = \frac{3}{4}\left(1 - \frac{x^{2}}{4}\right)$
$\displaystyle \Rightarrow \frac{9x^{2}}{16} + \frac{3x^{2}}{16} = \frac{3}{4}$
$\displaystyle \Rightarrow \frac{12x^{2}}{16} = \frac{3}{4} \Rightarrow x^{2} = 1$
$\displaystyle \Rightarrow x = \pm 1$
$\displaystyle \text{But } x = -1 \text{ does not satisfy the given equation.}$
$\displaystyle \text{Hence, } x = 1 \text{ is the required solution.}$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi,\text{ show that} \ \$
$\displaystyle x^{2}-y^{2}-z^{2}+2yz\sqrt{1-x^{2}}=0.\text{ \hspace{3.2cm} [ISC 2024]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi$
$\displaystyle \Rightarrow \sin^{-1}x + \sin^{-1}y = \pi - \sin^{-1}z$
$\displaystyle \Rightarrow \sin^{-1}(x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}}) = \pi - \sin^{-1}z$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}} = \sin(\pi - \sin^{-1}z)$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}} = \sin\pi\cos(\sin^{-1}z) - \cos\pi\sin(\sin^{-1}z)$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}} = 0 + z$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} = z - y\sqrt{1-x^{2}}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle x^{2}(1-y^{2}) = z^{2} + y^{2}(1-x^{2}) - 2yz\sqrt{1-x^{2}}$
$\displaystyle \Rightarrow x^{2} - x^{2}y^{2} = z^{2} + y^{2} - x^{2}y^{2} - 2yz\sqrt{1-x^{2}}$
$\displaystyle \Rightarrow x^{2} - y^{2} - z^{2} + 2yz\sqrt{1-x^{2}} = 0$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 4. }\text{Solve for }x:\ 5\tan^{-1}x+3\cot^{-1}x=2\pi.\text{ \hspace{3.2cm} [ISC 2023]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } 5\tan^{-1}x + 3\cot^{-1}x = 2\pi$
$\displaystyle \Rightarrow 2\tan^{-1}x + 3(\tan^{-1}x + \cot^{-1}x) = 2\pi$
$\displaystyle \Rightarrow 2\tan^{-1}x + 3\left(\frac{\pi}{2}\right) = 2\pi$
$\displaystyle \Rightarrow 2\tan^{-1}x = 2\pi - \frac{3\pi}{2}$
$\displaystyle \Rightarrow 2\tan^{-1}x = \frac{\pi}{2}$
$\displaystyle \Rightarrow \tan^{-1}x = \frac{\pi}{4}$
$\displaystyle \Rightarrow x = \tan\frac{\pi}{4} = 1$
$\\$

$\displaystyle \textbf{Question 5. }\text{If }\tan^{-1}\left(\frac{x-1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{2x+1}\right)=\tan^{-1}\left(\frac{23}{36}\right),\text{ then} \ \$
$\displaystyle \text{prove that }24x^{2}-23x-12=0.\text{ \hspace{3.2cm} [ISC 2023]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$
$\displaystyle \Rightarrow \tan^{-1}\left[\frac{\frac{x-1}{x+1} + \frac{2x-1}{2x+1}}{1 - \frac{(x-1)(2x-1)}{(x+1)(2x+1)}}\right] = \tan^{-1}\left(\frac{23}{36}\right)$
$\displaystyle \Rightarrow \text{Since } \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$
$\displaystyle \Rightarrow \frac{(x-1)(2x+1) + (2x-1)(x+1)}{(x+1)(2x+1) - (x-1)(2x-1)} = \frac{23}{36}$
$\displaystyle \Rightarrow \frac{2x^{2} - x - 1 + 2x^{2} + x - 1}{2x^{2} + 3x + 1 - 2x^{2} + 3x - 1} = \frac{23}{36}$
$\displaystyle \Rightarrow \frac{4x^{2} - 2}{6x} = \frac{23}{36}$
$\displaystyle \Rightarrow \frac{2(2x^{2} - 1)}{6x} = \frac{23}{36}$
$\displaystyle \Rightarrow 24x^{2} - 23x - 12 = 0$
$\\$

$\displaystyle \textbf{Question 6. } \text{Show that }\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\frac{\pi}{4}-\frac{x}{2},\ -\frac{\pi}{2}<x<\frac{\pi}{2}.\ \ \text{\hspace{0.2cm} [ISC 2022]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS }=\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$\displaystyle =\tan^{-1}\left(\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}+\sin^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}\right)$
$\displaystyle \left[\because \cos2A=\cos^{2}A-\sin^{2}A,\ \cos^{2}A+\sin^{2}A=1,\ \sin2A=2\sin A\cos A\right]$
$\displaystyle =\tan^{-1}\left(\frac{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}\right)$
$\displaystyle \left[\because a^{2}-b^{2}=(a+b)(a-b),\ (a^{2}+b^{2}+2ab)=(a+b)^{2}\right]$
$\displaystyle =\tan^{-1}\left(\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}\right)$
$\displaystyle =\tan^{-1}\left(\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}\right)$
$\displaystyle \left[\text{dividing numerator and denominator by }\cos\frac{x}{2}\right]$
$\displaystyle =\tan^{-1}\left[\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]$
$\displaystyle \left[\because \tan\left(\frac{\pi}{4}-\theta\right)=\frac{1-\tan\theta}{1+\tan\theta}\right]$
$\displaystyle =\frac{\pi}{4}-\frac{x}{2}\ \ \left[\because \tan^{-1}(\tan\theta)=\theta\right]\ \ \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 7. } \text{Prove that }\tan^{2}(\sec^{-1}2)+\cot^{2}(\mathrm{cosec}^{-1}3)=11\text{. \hspace{3.2cm} [ISC 2020]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \text{LHS}=\tan^{2}(\sec^{-1}2)+\cot^{2}(\mathrm{cosec}^{-1}3)$
$\displaystyle =\{\sec(\sec^{-1}2)\}^{2}-1+\{\mathrm{cosec}(\mathrm{cosec}^{-1}3)\}^{2}-1$
$\displaystyle [\because \tan^{2}\theta=\sec^{2}\theta-1 \text{ and } \cot^{2}\theta=\mathrm{cosec}^{2}\theta-1]$
$\displaystyle =(2)^{2}-1+(3)^{2}-1$
$\displaystyle =4-1+9-1=11=\text{RHS}\ \ \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 8. }\text{Evaluate }\cos\left(2\cos^{-1}x+\sin^{-1}x\right)\text{ at }x=\frac{1}{5}.\text{ \hspace{3.2cm} [ISC 2020]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \cos(2\cos^{-1}x + \sin^{-1}x)$
$\displaystyle = \cos(\cos^{-1}x + \cos^{-1}x + \sin^{-1}x)$
$\displaystyle = \cos\left(\cos^{-1}x + \frac{\pi}{2}\right)$
$\displaystyle \text{Since } \cos^{-1}\theta + \sin^{-1}\theta = \frac{\pi}{2}$
$\displaystyle = \cos\left(\frac{\pi}{2} + \cos^{-1}x\right)$
$\displaystyle = -\sin(\cos^{-1}x)$
$\displaystyle = -\sin(\sin^{-1}\sqrt{1-x^{2}})$
$\displaystyle = -\sqrt{1-x^{2}}$
$\displaystyle = -\sqrt{1 - \left(\frac{1}{5}\right)^{2}}$
$\displaystyle = -\sqrt{1 - \frac{1}{25}}$
$\displaystyle = -\sqrt{\frac{24}{25}}$
$\displaystyle = -\frac{2\sqrt{6}}{5}$
$\\$

$\displaystyle \textbf{Question 9. } \text{Solve for }x: \ \sin\left(2\tan^{-1}x\right)=1 \text{\hspace{3.2cm} [ISC 2019, 11]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \sin(2\tan^{-1}x)=1$
$\displaystyle \Rightarrow \sin(2\tan^{-1}x)=\sin\frac{\pi}{2}$
$\displaystyle \Rightarrow 2\tan^{-1}x=\frac{\pi}{2} \Rightarrow \tan^{-1}x=\frac{\pi}{4}$
$\displaystyle \therefore x=\tan\frac{\pi}{4}=1$
$\\$

$\displaystyle \textbf{Question 10. } \text{If }\sec^{-1}x=\mathrm{cosec}^{-1}y\text{, then show that }\frac{1}{x^{2}}+\frac{1}{y^{2}}=1.\text{ \hspace{0.2cm} [ISC 2019]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \sec^{-1}x=\mathrm{cosec}^{-1}y \quad ...(1)$
$\displaystyle \text{To show } \frac{1}{x^{2}}+\frac{1}{y^{2}}=1$
$\displaystyle \text{Let } \mathrm{cosec}^{-1}y=\theta \Rightarrow \mathrm{cosec}\theta=y$
$\displaystyle \text{In right angled }\triangle ABC,\ \text{using Pythagoras theorem, } AB=\sqrt{y^{2}-1}$
$\displaystyle \therefore \sec\theta=\frac{y}{\sqrt{y^{2}-1}}$
$\displaystyle \Rightarrow \theta=\sec^{-1}\left(\frac{y}{\sqrt{y^{2}-1}}\right)$
$\displaystyle \Rightarrow \mathrm{cosec}^{-1}y=\sec^{-1}\left(\frac{y}{\sqrt{y^{2}-1}}\right)$
$\displaystyle \therefore \text{From (1), } \sec^{-1}x=\sec^{-1}\left(\frac{y}{\sqrt{y^{2}-1}}\right)$
$\displaystyle \Rightarrow x=\frac{y}{\sqrt{y^{2}-1}} \Rightarrow x\sqrt{y^{2}-1}=y$
$\displaystyle \Rightarrow x^{2}(y^{2}-1)=y^{2}$
$\displaystyle \Rightarrow x^{2}y^{2}-x^{2}=y^{2}$
$\displaystyle \Rightarrow x^{2}y^{2}=x^{2}+y^{2}$
$\displaystyle \Rightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}=1\ \ \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 11. }\text{Solve }3\tan^{-1}x+\cot^{-1}x=\pi.\text{ \hspace{3.2cm} [ISC 2018]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } 3\tan^{-1}x + \cot^{-1}x = \pi$
$\displaystyle \Rightarrow 3\tan^{-1}x + \frac{\pi}{2} - \tan^{-1}x = \pi$
$\displaystyle \Rightarrow 2\tan^{-1}x = \pi - \frac{\pi}{2}$
$\displaystyle \Rightarrow 2\tan^{-1}x = \frac{\pi}{2}$
$\displaystyle \Rightarrow \tan^{-1}x = \frac{\pi}{4}$
$\displaystyle \Rightarrow x = \tan\frac{\pi}{4} \Rightarrow x = 1$
$\displaystyle \text{Hence, the value of } x \text{ is } 1.$
$\\$

$\displaystyle \textbf{Question 12. }\text{If }\tan^{-1}a+\tan^{-1}b+\tan^{-1}c=\pi,\text{ prove that } \\ a+b+c=abc.\text{ \hspace{3.2cm} [ISC 2018]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \tan^{-1}a + \tan^{-1}b + \tan^{-1}c = \pi$
$\displaystyle \text{Let } \tan^{-1}a = x \Rightarrow a = \tan x$
$\displaystyle \tan^{-1}b = y \Rightarrow b = \tan y$
$\displaystyle \tan^{-1}c = z \Rightarrow c = \tan z$
$\displaystyle \Rightarrow x + y + z = \pi \Rightarrow x + y = \pi - z$
$\displaystyle \tan(x + y) = \tan(\pi - z)$
$\displaystyle \Rightarrow \frac{\tan x + \tan y}{1 - \tan x \tan y} = -\tan z$
$\displaystyle \Rightarrow \frac{a + b}{1 - ab} = -c$
$\displaystyle \Rightarrow a + b = -c + abc$
$\displaystyle \Rightarrow a + b + c = abc$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 13. }\text{Prove that }\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)=\tan^{-1}\sqrt{x}.\text{ \hspace{3.2cm} [ISC 2017]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\cos^{-1}\left(\frac{1-(\sqrt{x})^{2}}{1+(\sqrt{x})^{2}}\right)$
$\displaystyle = \frac{1}{2}\times 2\tan^{-1}(\sqrt{x})$
$\displaystyle \text{Since } 2\tan^{-1}A = \cos^{-1}\left(\frac{1-A^{2}}{1+A^{2}}\right), \ A \geq 0$
$\displaystyle = \tan^{-1}(\sqrt{x}) = \text{RHS}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 14. }\text{Solve for }x,\ \sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x.\text{ \hspace{0.2cm} [ISC 2017, 13]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \sin^{-1}x + \sin^{-1}(1-x) = \cos^{-1}x$
$\displaystyle \therefore \sin^{-1}\left[x\sqrt{1-(1-x)^{2}} + (1-x)\sqrt{1-x^{2}}\right] = \cos^{-1}x$
$\displaystyle \text{Since } \sin^{-1}A + \sin^{-1}B = \sin^{-1}\left(A\sqrt{1-B^{2}} + B\sqrt{1-A^{2}}\right)$
$\displaystyle \Rightarrow \sin^{-1}\left[x\sqrt{1-(1+x^{2}-2x)} + (1-x)\sqrt{1-x^{2}}\right] = \sin^{-1}\sqrt{1-x^{2}}$
$\displaystyle \text{Since } \cos^{-1}A = \sin^{-1}\sqrt{1-A^{2}}$
$\displaystyle \Rightarrow x\sqrt{2x-x^{2}} + (1-x)\sqrt{1-x^{2}} = \sqrt{1-x^{2}}$
$\displaystyle \Rightarrow x\sqrt{2x-x^{2}} = (1-1+x)\sqrt{1-x^{2}}$
$\displaystyle \Rightarrow x\sqrt{2x-x^{2}} = x\sqrt{1-x^{2}}$
$\displaystyle \Rightarrow x\sqrt{2x-x^{2}} - x\sqrt{1-x^{2}} = 0$
$\displaystyle \Rightarrow x\left[\sqrt{2x-x^{2}} - \sqrt{1-x^{2}}\right] = 0$
$\displaystyle \Rightarrow x = 0 \text{ or } \sqrt{2x-x^{2}} - \sqrt{1-x^{2}} = 0$
$\displaystyle \text{Now, } \sqrt{2x-x^{2}} - \sqrt{1-x^{2}} = 0$
$\displaystyle \Rightarrow \sqrt{2x-x^{2}} = \sqrt{1-x^{2}}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle 2x-x^{2} = 1-x^{2} \Rightarrow 2x = 1$
$\displaystyle \therefore x = \frac{1}{2}$
$\displaystyle \text{Hence, } x = 0, \frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 15. }\text{Solve for }x,\ \text{if }\tan(\cos^{-1}x)=\frac{2}{\sqrt{5}}.\text{ \hspace{3.2cm} [ISC 2016]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \tan(\cos^{-1}x) = \frac{2}{\sqrt{5}}$
$\displaystyle \Rightarrow \tan\left(\tan^{-1}\frac{\sqrt{1-x^{2}}}{x}\right) = \frac{2}{\sqrt{5}}$
$\displaystyle \text{Since } \cos^{-1}A = \tan^{-1}\left(\frac{\sqrt{1-A^{2}}}{A}\right)$
$\displaystyle \Rightarrow \frac{\sqrt{1-x^{2}}}{x} = \frac{2}{\sqrt{5}}$
$\displaystyle \text{Since } \tan(\tan^{-1}\theta) = \theta$
$\displaystyle \Rightarrow \sqrt{5}\sqrt{1-x^{2}} = 2x$
$\displaystyle \Rightarrow 5(1-x^{2}) = 4x^{2}$
$\displaystyle \Rightarrow 5 - 5x^{2} = 4x^{2}$
$\displaystyle \Rightarrow 9x^{2} - 5 = 0$
$\displaystyle \Rightarrow x^{2} = \frac{5}{9} \Rightarrow x = \pm \frac{\sqrt{5}}{3}$
$\\$

$\displaystyle \textbf{Question 16. }\text{If }\sin^{-1}x+\tan^{-1}x=\frac{\pi}{2},\text{ then prove that }2x^{2}+1=\sqrt{5}.\text{ \hspace{0.2cm} [ISC 2016]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \sin^{-1}x + \tan^{-1}x = \frac{\pi}{2}$
$\displaystyle \Rightarrow \tan^{-1}x = \left(\frac{\pi}{2} - \sin^{-1}x\right)$
$\displaystyle \Rightarrow \tan^{-1}x = \cos^{-1}x$
$\displaystyle \text{Since } \cos^{-1}\theta + \sin^{-1}\theta = \frac{\pi}{2}$
$\displaystyle \Rightarrow \tan^{-1}x = \tan^{-1}\frac{\sqrt{1-x^{2}}}{x}$
$\displaystyle \text{Since } \cos^{-1}\theta = \tan^{-1}\frac{\sqrt{1-\theta^{2}}}{\theta}$
$\displaystyle \Rightarrow x = \tan\left(\tan^{-1}\frac{\sqrt{1-x^{2}}}{x}\right)$
$\displaystyle \Rightarrow x = \frac{\sqrt{1-x^{2}}}{x}$
$\displaystyle \Rightarrow x^{2} = \sqrt{1-x^{2}}$
$\displaystyle \Rightarrow (x^{2})^{2} = 1-x^{2}$
$\displaystyle \Rightarrow x^{4} + x^{2} - 1 = 0$
$\displaystyle \therefore x^{2} = \frac{-1 \pm \sqrt{5}}{2}$
$\displaystyle \text{Using } x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$
$\displaystyle \Rightarrow 2x^{2} + 1 = \pm \sqrt{5}$
$\displaystyle \text{Since } 2x^{2} + 1 > 0, \ 2x^{2} + 1 = -\sqrt{5} \text{ is not possible}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 17. }\text{Solve for }x:\ \cos^{-1}\left(\sin(\cos^{-1}x)\right)=\frac{\pi}{6}.\text{ \hspace{3.2cm} [ISC 2015]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \cos^{-1}(\sin(\cos^{-1}x)) = \frac{\pi}{6}$
$\displaystyle \Rightarrow \sin(\cos^{-1}x) = \cos\frac{\pi}{6}$
$\displaystyle \Rightarrow \sin\left(\sin^{-1}\sqrt{1-x^{2}}\right) = \frac{\sqrt{3}}{2}$
$\displaystyle \text{Since } \cos^{-1}A = \sin^{-1}\sqrt{1-A^{2}}$
$\displaystyle \Rightarrow \sqrt{1-x^{2}} = \frac{\sqrt{3}}{2}$
$\displaystyle \text{Since } \sin(\sin^{-1}\theta) = \theta$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle 1 - x^{2} = \frac{3}{4} \Rightarrow x^{2} = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 18. }\text{Solve the equation for }x:\ \sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}=\frac{\pi}{2}, \\ \ x\neq 0.\text{ \hspace{8.2cm} [ISC 2015]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \sin^{-1}\frac{5}{x} + \sin^{-1}\frac{12}{x} = \frac{\pi}{2}$
$\displaystyle \text{Now, } \sin^{-1}\frac{5}{x} + \cos^{-1}\sqrt{1-\left(\frac{12}{x}\right)^{2}} = \frac{\pi}{2} \qquad \ldots (i)$
$\displaystyle \text{Since } \sin^{-1}\theta = \cos^{-1}\sqrt{1-\theta^{2}}$
$\displaystyle \text{We know that } \sin^{-1}\theta + \cos^{-1}\theta = \frac{\pi}{2}$
$\displaystyle \therefore \frac{5}{x} = \sqrt{1-\frac{144}{x^{2}}}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \frac{25}{x^{2}} = 1-\frac{144}{x^{2}}$
$\displaystyle \Rightarrow \frac{25+144}{x^{2}} = 1$
$\displaystyle \Rightarrow \frac{169}{x^{2}} = 1$
$\displaystyle \Rightarrow x^{2} = 169 \Rightarrow x = \pm 13$
$\displaystyle \text{But } x = -13 \text{ does not satisfy the given equation.}$
$\displaystyle \text{Hence, required value of } x \text{ is } 13.$
$\\$

$\displaystyle \textbf{Question 19. }\text{Evaluate }\tan\left[2\tan^{-1}\frac{1}{2}-\cot^{-1}3\right].\text{ \hspace{3.2cm} [ISC 2014]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \tan\left[2\tan^{-1}\frac{1}{2} - \cot^{-1}3\right]$
$\displaystyle = \tan\left[\tan^{-1}\left(\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^{2}}\right) - \tan^{-1}\frac{1}{3}\right]$
$\displaystyle \text{Since } 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^{2}}\right) \text{ and } \cot^{-1}x = \tan^{-1}\frac{1}{x}$
$\displaystyle = \tan\left[\tan^{-1}\frac{1}{\frac{3}{4}} - \tan^{-1}\frac{1}{3}\right]$
$\displaystyle = \tan\left[\tan^{-1}\frac{4}{3} - \tan^{-1}\frac{1}{3}\right]$
$\displaystyle = \tan\left[\tan^{-1}\left(\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{3}\cdot\frac{1}{3}}\right)\right]$
$\displaystyle \text{Since } \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$
$\displaystyle = \tan\left[\tan^{-1}\left(\frac{\frac{3}{3}}{\frac{9+4}{9}}\right)\right]$
$\displaystyle = \frac{3}{1}\cdot\frac{9}{13} = \frac{9}{13}$
$\displaystyle \text{Since } \tan(\tan^{-1}x) = x$
$\\$

$\displaystyle \textbf{Question 20. }\text{If }\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}=\theta,\text{ then prove that} \ \$
$\displaystyle 9x^{2}-12xy\cos\theta+4y^{2}=36\sin^{2}\theta.\text{ \hspace{5.2cm} [ISC 2014]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \cos^{-1}\frac{x}{2} + \cos^{-1}\frac{y}{3} = \theta$
$\displaystyle \Rightarrow \cos^{-1}\left[\left(\frac{x}{2}\cdot\frac{y}{3}\right) - \sqrt{1-\frac{x^{2}}{4}}\sqrt{1-\frac{y^{2}}{9}}\right] = \theta$
$\displaystyle \text{Since } \cos^{-1}u + \cos^{-1}v = \cos^{-1}\left(uv-\sqrt{(1-u^{2})(1-v^{2})}\right)$
$\displaystyle \Rightarrow \frac{xy}{6} - \sqrt{\left(1-\frac{x^{2}}{4}\right)\left(1-\frac{y^{2}}{9}\right)} = \cos\theta$
$\displaystyle \Rightarrow \frac{xy}{6} - \cos\theta = \sqrt{\left(1-\frac{x^{2}}{4}\right)\left(1-\frac{y^{2}}{9}\right)}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \frac{x^{2}y^{2}}{36} - \frac{xy\cos\theta}{3} + \cos^{2}\theta = \left(1-\frac{x^{2}}{4}\right)\left(1-\frac{y^{2}}{9}\right)$
$\displaystyle \Rightarrow \frac{x^{2}y^{2}}{36} - \frac{xy\cos\theta}{3} + \cos^{2}\theta = 1-\frac{x^{2}}{4}-\frac{y^{2}}{9}+\frac{x^{2}y^{2}}{36}$
$\displaystyle \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{9} - \frac{xy\cos\theta}{3} = 1-\cos^{2}\theta$
$\displaystyle \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{9} - \frac{xy\cos\theta}{3} = \sin^{2}\theta$
$\displaystyle \Rightarrow 9x^{2} + 4y^{2} - 12xy\cos\theta = 36\sin^{2}\theta$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 21. }\text{If }\cos^{-1}x+\cos^{-1}y+\cos^{-1}z=\pi,\text{ then prove that} \ \$
$\displaystyle x^{2}+y^{2}+z^{2}+2xyz=1.\text{ \hspace{5.2cm} [ISC 2014]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi$
$\displaystyle \Rightarrow \cos^{-1}x + \cos^{-1}y = \pi - \cos^{-1}z$
$\displaystyle \Rightarrow \cos^{-1}\left(xy - \sqrt{1-x^{2}}\sqrt{1-y^{2}}\right) = \pi - \cos^{-1}z$
$\displaystyle \text{Since } \cos^{-1}A + \cos^{-1}B = \cos^{-1}(AB - \sqrt{1-A^{2}}\sqrt{1-B^{2}})$
$\displaystyle \Rightarrow xy - \sqrt{1-x^{2}}\sqrt{1-y^{2}} = \cos(\pi - \cos^{-1}z)$
$\displaystyle \Rightarrow xy - \sqrt{1-x^{2}}\sqrt{1-y^{2}} = -\cos(\cos^{-1}z)$
$\displaystyle \Rightarrow xy - \sqrt{1-x^{2}}\sqrt{1-y^{2}} = -z$
$\displaystyle \Rightarrow \sqrt{1-x^{2}}\sqrt{1-y^{2}} = xy + z$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle (1-x^{2})(1-y^{2}) = (xy+z)^{2}$
$\displaystyle \Rightarrow 1 - x^{2} - y^{2} + x^{2}y^{2} = x^{2}y^{2} + z^{2} + 2xyz$
$\displaystyle \Rightarrow x^{2} + y^{2} + z^{2} + 2xyz = 1$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Prove that }\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\frac{1}{2}\sin^{-1}\frac{4}{5}.\text{ \hspace{3.2cm} [ISC 2013]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9}$
$\displaystyle = \tan^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\cdot\frac{2}{9}}\right) = \tan^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right)$
$\displaystyle \text{Since } \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$
$\displaystyle = \tan^{-1}\left(\frac{17}{34}\right) = \tan^{-1}\frac{1}{2} \qquad \ldots (i)$
$\displaystyle = \frac{1}{2}\sin^{-1}\left(\frac{2\times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^{2}}\right)$
$\displaystyle \text{Since } 2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)$
$\displaystyle = \frac{1}{2}\sin^{-1}\left(\frac{1}{1+\frac{1}{4}}\right) = \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right) = \text{RHS}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 23. }\text{Prove that }\cos^{-1}\frac{63}{65}+2\tan^{-1}\frac{1}{5}=\sin^{-1}\frac{3}{5}.\text{ \hspace{2.2cm} [ISC 2012]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = \cos^{-1}\left(\frac{63}{65}\right) + 2\tan^{-1}\left(\frac{1}{5}\right)$
$\displaystyle = \sin^{-1}\sqrt{1-\left(\frac{63}{65}\right)^{2}} + \sin^{-1}\left(\frac{2\times \frac{1}{5}}{1+\left(\frac{1}{5}\right)^{2}}\right)$
$\displaystyle \text{Since } \cos^{-1}x = \sin^{-1}\sqrt{1-x^{2}} \text{ and } 2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)$
$\displaystyle = \sin^{-1}\sqrt{\frac{65^{2}-63^{2}}{65^{2}}} + \sin^{-1}\left(\frac{2/5}{26/25}\right)$
$\displaystyle = \sin^{-1}\sqrt{\frac{(65-63)(65+63)}{65^{2}}} + \sin^{-1}\left(\frac{2}{5}\cdot\frac{25}{26}\right)$
$\displaystyle = \sin^{-1}\sqrt{\frac{2\times 128}{65^{2}}} + \sin^{-1}\left(\frac{5}{13}\right)$
$\displaystyle = \sin^{-1}\left(\frac{16}{65}\right) + \sin^{-1}\left(\frac{5}{13}\right)$
$\displaystyle = \sin^{-1}\left[\frac{16}{65}\sqrt{1-\left(\frac{5}{13}\right)^{2}} + \frac{5}{13}\sqrt{1-\left(\frac{16}{65}\right)^{2}}\right]$
$\displaystyle \text{Since } \sin^{-1}A + \sin^{-1}B = \sin^{-1}\left(A\sqrt{1-B^{2}} + B\sqrt{1-A^{2}}\right)$
$\displaystyle = \sin^{-1}\left[\frac{16}{65}\cdot\frac{12}{13} + \frac{5}{13}\cdot\frac{63}{65}\right]$
$\displaystyle = \sin^{-1}\left(\frac{192 + 315}{65\times 13}\right)$
$\displaystyle = \sin^{-1}\left(\frac{507}{845}\right) = \sin^{-1}\left(\frac{3}{5}\right)$
$\displaystyle = \text{RHS}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 24. } \text{Prove that }\sec^{2}(\tan^{-1}2)+\mathrm{cosec}^{2}(\cot^{-1}3)=15.\text{ \hspace{0.2cm} [ISC 2012]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sec^{2}(\tan^{-1}2)+\mathrm{cosec}^{2}(\cot^{-1}3)$
$\displaystyle \text{Let } \tan^{-1}2=\theta \text{ and } \cot^{-1}3=\phi$
$\displaystyle \Rightarrow \tan\theta=2 \text{ and } \cot\phi=3$
$\displaystyle \text{Then, LHS}=\sec^{2}\theta+\mathrm{cosec}^{2}\phi$
$\displaystyle =(1+\tan^{2}\theta)+(1+\cot^{2}\phi)$
$\displaystyle [\because \sec^{2}A=1+\tan^{2}A \text{ and } \mathrm{cosec}^{2}A=1+\cot^{2}A]$
$\displaystyle =(1+4)+(1+9)=5+10=15=\text{RHS}\ \ \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 25. }\text{Solve for }x:\ \tan^{-1}\left(\frac{x-1}{x-2}\right)+\tan^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}.\text{ \hspace{0.2cm} [ISC 2011]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4}$
$\displaystyle \Rightarrow \tan^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2}\cdot\frac{x+1}{x+2}}\right] = \frac{\pi}{4}, \ \frac{(x-1)(x+1)}{(x-2)(x+2)} < 1$
$\displaystyle \text{Since } \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), \text{ if } xy < 1$
$\displaystyle \Rightarrow \tan^{-1}\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}\right] = \frac{\pi}{4}, \ \frac{x^{2}-1}{x^{2}-4} < 1$
$\displaystyle \Rightarrow \frac{(x^{2}+x-2)+(x^{2}-x-2)}{x^{2}-4-(x^{2}-1)} = \tan\frac{\pi}{4}, \ \frac{3}{x^{2}-4} < 0$
$\displaystyle \Rightarrow \frac{2x^{2}-4}{-3} = 1, \ \frac{3}{x^{2}-4} < 0$
$\displaystyle \Rightarrow 2x^{2}-4 = -3, \ x^{2}-4 < 0$
$\displaystyle \Rightarrow 2x^{2} = 1, \ x^{2} < 4$
$\displaystyle \Rightarrow x = \pm \frac{1}{\sqrt{2}}, \ -2 < x < 2$
$\displaystyle \text{Hence, both values of } x \text{ satisfy the given equation.}$
$\\$

$\displaystyle \textbf{Question 26. }\text{Prove that }2\tan^{-1}\frac{1}{5}+\cos^{-1}\frac{7}{5\sqrt{2}}+2\tan^{-1}\frac{1}{8}=\frac{\pi}{4}.\text{ \hspace{0.2cm} [ISC 2011]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = 2\tan^{-1}\frac{1}{5} + \cos^{-1}\frac{7}{5\sqrt{2}} + 2\tan^{-1}\frac{1}{8}$
$\displaystyle = \tan^{-1}\left(\frac{2\times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right) + \tan^{-1}\left(\frac{\sqrt{1-\frac{49}{50}}}{\frac{7}{5\sqrt{2}}}\right) + \tan^{-1}\left(\frac{2\times \frac{1}{8}}{1-\left(\frac{1}{8}\right)^{2}}\right)$
$\displaystyle \text{Since } 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^{2}}\right), \ \cos^{-1}x = \tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$
$\displaystyle = \tan^{-1}\left(\frac{2/5}{24/25}\right) + \tan^{-1}\left(\frac{1/5\sqrt{2}}{7/5\sqrt{2}}\right) + \tan^{-1}\left(\frac{2/8}{63/64}\right)$
$\displaystyle = \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{16}{63}\right)$
$\displaystyle = \tan^{-1}\left(\frac{\frac{1}{7}+\frac{16}{63}}{1-\frac{1}{7}\cdot\frac{16}{63}}\right) + \tan^{-1}\frac{5}{12}$
$\displaystyle \text{Since } \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$
$\displaystyle = \tan^{-1}\left(\frac{63+112}{441-16}\right) + \tan^{-1}\frac{5}{12}$
$\displaystyle = \tan^{-1}\left(\frac{175}{425}\right) + \tan^{-1}\frac{5}{12} = \tan^{-1}\frac{7}{17} + \tan^{-1}\frac{5}{12}$
$\displaystyle = \tan^{-1}\left(\frac{\frac{7}{17}+\frac{5}{12}}{1-\frac{7}{17}\cdot\frac{5}{12}}\right)$
$\displaystyle = \tan^{-1}\left(\frac{84+85}{204-35}\right) = \tan^{-1}(1)$
$\displaystyle = \frac{\pi}{4} = \text{RHS}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 27. }\text{Prove that }\sin\left[2\tan^{-1}\frac{3}{5}-\sin^{-1}\frac{7}{25}\right]=\frac{304}{425}.\text{ \hspace{2.2cm} [ISC 2010]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ To prove, } \sin\left[2\tan^{-1}\frac{3}{5} - \sin^{-1}\frac{7}{25}\right] = \frac{304}{425}$
$\displaystyle \text{i.e. } 2\tan^{-1}\frac{3}{5} - \sin^{-1}\frac{7}{25} = \sin^{-1}\frac{304}{425}$
$\displaystyle \text{LHS } = 2\tan^{-1}\frac{3}{5} - \sin^{-1}\frac{7}{25}$
$\displaystyle = \tan^{-1}\frac{3}{5} + \tan^{-1}\frac{3}{5} - \sin^{-1}\frac{7}{25}$
$\displaystyle = \sin^{-1}\left(\frac{\frac{3}{5}}{\sqrt{1+\frac{9}{25}}}\right) + \sin^{-1}\left(\frac{\frac{3}{5}}{\sqrt{1+\frac{9}{25}}}\right) - \sin^{-1}\frac{7}{25}$
$\displaystyle \text{Since } \tan^{-1}x = \sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$
$\displaystyle = \sin^{-1}\frac{3}{\sqrt{34}} + \sin^{-1}\frac{3}{\sqrt{34}} - \sin^{-1}\frac{7}{25}$
$\displaystyle = \sin^{-1}\left[\frac{3}{\sqrt{34}}\sqrt{1-\frac{9}{34}} + \frac{3}{\sqrt{34}}\sqrt{1-\frac{9}{34}}\right] - \sin^{-1}\frac{7}{25}$
$\displaystyle \text{Since } \sin^{-1}x + \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}}\right)$
$\displaystyle = \sin^{-1}\left(\frac{3}{\sqrt{34}}\cdot\frac{5}{\sqrt{34}} + \frac{3}{\sqrt{34}}\cdot\frac{5}{\sqrt{34}}\right) - \sin^{-1}\frac{7}{25}$
$\displaystyle = \sin^{-1}\frac{30}{34} - \sin^{-1}\frac{7}{25}$
$\displaystyle = \sin^{-1}\left[\frac{30}{34}\sqrt{1-\frac{49}{625}} - \frac{7}{25}\sqrt{1-\frac{900}{1156}}\right]$
$\displaystyle \text{Since } \sin^{-1}x - \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^{2}} - y\sqrt{1-x^{2}}\right)$
$\displaystyle = \sin^{-1}\left[\frac{30}{34}\sqrt{\frac{625-49}{625}} - \frac{7}{25}\sqrt{\frac{1156-900}{1156}}\right]$
$\displaystyle = \sin^{-1}\left(\frac{30}{34}\cdot\frac{24}{25} - \frac{7}{25}\cdot\frac{16}{34}\right)$
$\displaystyle = \sin^{-1}\left(\frac{720-112}{850}\right)$
$\displaystyle = \sin^{-1}\frac{608}{850} = \sin^{-1}\frac{304}{425} = \text{RHS}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi,\text{ then prove that} \ \$
$\displaystyle x^{2}-y^{2}-z^{2}+2yz\sqrt{1-x^{2}}=0.\text{ \hspace{5.2cm} [ISC 2009]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi$
$\displaystyle \Rightarrow \sin^{-1}x + \sin^{-1}y = \pi - \sin^{-1}z$
$\displaystyle \Rightarrow \sin^{-1}\left[x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}}\right] = \pi - \sin^{-1}z$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}} = \sin(\pi - \sin^{-1}z)$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}} = \sin(\sin^{-1}z)$
$\displaystyle \text{Since } \sin(\pi-\theta) = \sin\theta$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} + y\sqrt{1-x^{2}} = z$
$\displaystyle \Rightarrow x\sqrt{1-y^{2}} = z - y\sqrt{1-x^{2}}$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle x^{2}(1-y^{2}) = z^{2} + y^{2}(1-x^{2}) - 2zy\sqrt{1-x^{2}}$
$\displaystyle \Rightarrow x^{2} - x^{2}y^{2} = z^{2} + y^{2} - x^{2}y^{2} - 2zy\sqrt{1-x^{2}}$
$\displaystyle \therefore x^{2} - y^{2} - z^{2} + 2zy\sqrt{1-x^{2}} = 0$
$\\$

$\displaystyle \textbf{Question 29. } \text{Prove that }\cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)=7.\text{ \hspace{3.2cm} [ISC 2005]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS }=\cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)$
$\displaystyle \text{Let } \cot^{-1}3=\theta$
$\displaystyle \Rightarrow \cot\theta=3$
$\displaystyle \Rightarrow \tan\theta=\frac{1}{3}$
$\displaystyle \therefore \cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)=\cot\left(\frac{\pi}{4}-2\theta\right)$
$\displaystyle =\frac{1}{\tan\left(\frac{\pi}{4}-2\theta\right)}$
$\displaystyle =\frac{1+\tan\frac{\pi}{4}\cdot\tan2\theta}{\tan\frac{\pi}{4}-\tan2\theta}$
$\displaystyle \left[\because \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right]$
$\displaystyle =\frac{1+\tan2\theta}{1-\tan2\theta}$
$\displaystyle =\frac{1+\frac{2\tan\theta}{1-\tan^{2}\theta}}{1-\frac{2\tan\theta}{1-\tan^{2}\theta}}$
$\displaystyle \left[\because \tan2A=\frac{2\tan A}{1-\tan^{2}A}\right]$
$\displaystyle =\frac{1+\frac{2\cdot\frac{1}{3}}{1-\frac{1}{9}}}{1-\frac{2\cdot\frac{1}{3}}{1-\frac{1}{9}}}$
$\displaystyle =\frac{1+\frac{2}{3}\cdot\frac{9}{8}}{1-\frac{2}{3}\cdot\frac{9}{8}}$
$\displaystyle =\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$
$\displaystyle =\frac{\frac{7}{4}}{\frac{1}{4}}=7=\text{RHS}\ \ \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 30. }\text{Prove that }2\left(\tan^{-1}1+\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}\right)=\pi.\text{ \hspace{0.2cm} [ISC 2004]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = 2\left[\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right]$
$\displaystyle = 2\left[\tan^{-1}\left(\frac{1+\frac{1}{2}}{1-1\cdot \frac{1}{2}}\right) + \tan^{-1}\frac{1}{3}\right]$
$\displaystyle \text{Since } \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$
$\displaystyle = 2\left[\tan^{-1}3 + \tan^{-1}\frac{1}{3}\right] = 2\tan^{-1}\left(\frac{3+\frac{1}{3}}{1-3\cdot \frac{1}{3}}\right)$
$\displaystyle = 2\tan^{-1}\left(\frac{\frac{10}{3}}{0}\right) = 2\tan^{-1}\infty$
$\displaystyle = 2\cdot \frac{\pi}{2} = \pi = \text{RHS}$
$\displaystyle \text{Since } \tan\frac{\pi}{2} = \infty$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 31. }\text{Show that }\sin^{-1}\frac{4}{5}+\cos^{-1}\frac{2}{\sqrt{5}}=\cot^{-1}\frac{2}{11}.\text{ \hspace{0.2cm} [ISC 2003]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = \sin^{-1}\frac{4}{5} + \cos^{-1}\frac{2}{\sqrt{5}}$
$\displaystyle = \tan^{-1}\left(\frac{\frac{4}{5}}{\sqrt{1-\frac{16}{25}}}\right) + \tan^{-1}\left(\frac{\sqrt{1-\frac{4}{5}}}{\frac{2}{\sqrt{5}}}\right)$
$\displaystyle \text{Since } \sin^{-1}x = \tan^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \text{ and } \cos^{-1}x = \tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$
$\displaystyle = \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{1}{2}$
$\displaystyle = \tan^{-1}\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3}\cdot \frac{1}{2}}\right)$
$\displaystyle \text{Since } \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$
$\displaystyle = \tan^{-1}\left(\frac{8+3}{6-4}\right) = \tan^{-1}\left(\frac{11}{2}\right)$
$\displaystyle = \cot^{-1}\frac{2}{11} = \text{RHS}$
$\displaystyle \text{Since } \tan^{-1}x = \cot^{-1}\frac{1}{x}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 32. }\text{Solve }\tan^{-1}(2+x)+\tan^{-1}(2-x)=\tan^{-1}\frac{2}{3}.\text{ \hspace{0.2cm} [ISC 2001]} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \tan^{-1}(2+x) + \tan^{-1}(2-x) = \tan^{-1}\left(\frac{2}{3}\right)$
$\displaystyle \Rightarrow \tan^{-1}\left[\frac{(2+x)+(2-x)}{1-(2+x)(2-x)}\right] = \tan^{-1}\left(\frac{2}{3}\right)$
$\displaystyle \text{Since } \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$
$\displaystyle \Rightarrow \tan^{-1}\left(\frac{4}{1-(4-x^{2})}\right) = \tan^{-1}\left(\frac{2}{3}\right)$
$\displaystyle \Rightarrow \tan^{-1}\left(\frac{4}{x^{2}-3}\right) = \tan^{-1}\left(\frac{2}{3}\right)$
$\displaystyle \Rightarrow \frac{4}{x^{2}-3} = \frac{2}{3}$
$\displaystyle \Rightarrow 12 = 2x^{2} - 6 \Rightarrow 2x^{2} = 18 \Rightarrow x^{2} = 9$
$\displaystyle \therefore x = \pm 3$
$\\$