Inverse Trigonometric Functions

$\begin{aligned} & \tan ^{-1}(-2)-\tan ^{-1}(3)=\tan ^{-1}\left[\frac{(-2)-3}{1+(-2)(3)}\right] \\ & =\tan ^{-1}\left[\frac{-5}{-5}\right]=\tan ^{-1} 1\end{aligned}$

From options, $\tan \frac{3 \pi}{4}=-1$

$\begin{aligned} & \Rightarrow \quad \tan \left(\frac{-3 \pi}{4}\right)=1 \Rightarrow \tan ^{-1}(1)=\frac{-3 \pi}{4} \\ & \therefore \tan ^{-1}(-2)-\tan ^{-1}(3)=\frac{-3 \pi}{4} \end{aligned}$

$\begin{aligned} & x=\sin \left(2 \tan ^{-1} 2\right)=\frac{2 \cdot 3}{1+3^2}=\frac{3}{5} \\ & y=\cos \left(2 \tan ^{-1} 3\right)=\frac{\left(1-3^2\right)}{\left(1+3^2\right)}=\frac{-4}{5} \\ & z=\sec \left(3 \tan ^{-1} 4\right)=\tan ^{-1} 4=\theta \\ & \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta \\ & =4 \cdot\left(\frac{1}{17 \sqrt{17}}\right)-3\left(\frac{1}{\sqrt{17}}\right) \\ & \end{aligned}$

$=\frac{-47}{17\sqrt{17}}$

$\sec3\theta=\frac{-17\sqrt{17}}{47}$

$\therefore \quad x > y > z$.

Let $x=\cos 2 \theta$

$\begin{aligned} & \Rightarrow \frac{d \theta}{d x}=-\frac{1}{\sin 2 \theta} \\ & \therefore \frac{d}{d x}\left\{\sin ^{-1}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\right\}=\frac{d}{d x} \sin ^2 \theta=\sin 2 \theta \frac{d \theta}{d x}=-1 \end{aligned}$

$\begin{aligned} y^{\prime} & =\left[\frac{1}{1+\left(\frac{a x-b}{b x+a}\right)^2}\right] \frac{(b x+a) a-(a x-b) b}{(b x+a)^2} \\ & =\left[\frac{1}{\left(a^2+b^2\right)\left(x^2+1\right)}\right]\left(a^2+b^2\right)=\frac{1}{x^2+1}\end{aligned}$

$\begin{aligned} & \text { } \sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0 \\ & \Rightarrow \quad 2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\sin ^{-1}(0)=0 \\ & \Rightarrow \quad \cos ^{-1}\left(\cot \left(2 \tan ^{-1} x\right)\right)=0 \\ & \Rightarrow \quad \cot \left[\left(2 \tan ^{-1} x\right)\right]=\cos 0=1 \\ & \Rightarrow \quad \cot \left[\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]=1 \\ & \Rightarrow \quad \cot \left[\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)\right]=1 \\ & \Rightarrow \quad \frac{1-x^2}{2 x}=1 \\ & \Rightarrow \quad 1-x^2=2 x \\ & \Rightarrow \quad x^2+2 x-1=0 \\ & x=\frac{-2 \pm \sqrt{4+4}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2} \\ & x=-1 \pm \sqrt{2} \\ \end{aligned}$

Given equation holds for 2 values of $x$.

$\begin{aligned} & \text { } \tan ^{-1}\left(\frac{1}{1+1 \cdot 2}\right)+\tan ^{-1}\left(\frac{1}{1+2 \cdot 3}\right)+\ldots \\ & \quad+\tan ^{-1}\left(\frac{1}{1+n(n+1)}\right)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{2-1}{1+1 \cdot 2}\right)+\tan ^{-1}\left(\frac{3-2}{1+2 \cdot 3}\right)+\ldots \\ & \quad+\tan ^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}(2)-\tan ^{-1}(1)+\tan ^{-1}(3)-\tan ^{-1}(2)+\ldots \\ & \Rightarrow \quad \tan ^{-1}(n+1)-\tan ^{-1}(n)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{n+1-1}{1+(n+1) \cdot 1}\right)=\tan ^{-1}(x)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{n}{n+2}\right)=\tan ^{-1}(x) \Rightarrow x=\frac{n}{n+2}\end{aligned}$

$y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$

Substituting $x^2=\cos 2 \theta$

$\begin{aligned} \Rightarrow \quad \theta & =\frac{1}{2} \cos ^{-1}\left(x^2\right) \\ \therefore \quad y & =\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right) \\ & =\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \Rightarrow y=\frac{\pi}{4}+\theta \\ \Rightarrow \quad y & =\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(x^2\right) \\ \therefore \quad \frac{d y}{d x} & =\frac{1}{2} \cdot\left(\frac{-2 x}{\sqrt{1-x^4}}\right)=\frac{-x}{\sqrt{1-x^4}}\end{aligned}$

$\frac{1}{k} \sec ^{-1}\left(x^k\right)=\int \frac{d x}{x \sqrt{x^4-1}}=\frac{1}{2} \int \frac{2 x d x}{x^2 \sqrt{x^4-1}}$

Let

$\begin{aligned} & \Rightarrow \quad 2 x d x=d t \\ & =\frac{1}{2} \int \frac{d t}{t \sqrt{t^2-1}}=\frac{1}{2} \sec ^{-1} t \\ & \Rightarrow \frac{1}{k} \sec ^{-1}\left(x^k\right)=\frac{1}{2} \sec ^{-1}\left(x^2\right) \Rightarrow k=2 \\ & \end{aligned}$

Let $I = {({\sin ^{ - 1}}x)^3} + {({\cos ^{ - 1}}x)^3}$

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

$ = {\pi \over 2}\left[ {{{({{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x)}^2} - 3{{\sin }^{ - 1}}x\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \right]$

$ = {\pi \over 2}\left( {{{{\pi ^2}} \over 4} - {{3\pi } \over 2}{{\sin }^{ - 1}}x + 3{{({{\sin }^{ - 1}}x)}^2}} \right)$

$ = {\pi \over 2}\left[ {3\left( {{{({{\sin }^{ - 1}}x)}^2} - {\pi \over 2}({{\sin }^1}x) + {{{\pi ^2}} \over {16}} - {{{\pi ^2}} \over {16}}} \right) + {{{\pi ^2}} \over 4}} \right]$

$ = {\pi \over 2}\left( {{{{\pi ^2}} \over {16}} + 3{{\left( {{{\sin }^{ - 1}}x - {\pi \over 4}} \right)}^2}} \right)$

Now, ${\pi \over 4} \ge {\sin ^{ - 1}}x - {\pi \over 4} \ge - {{3\pi } \over 4}$

$ \Rightarrow {{9{\pi ^2}} \over {16}} \ge {\left( {{{\sin }^{ - 1}}x - {\pi \over 4}} \right)^2} \ge 0 \Rightarrow I \ge {\pi \over 2}\,.\,{{{\pi ^2}} \over {16}} = {{{\pi ^3}} \over {32}}$

$\left(\sin ^{-1} y\right)^2+\cos ^{-1} x=\frac{n \pi^2}{4}$ and $\left(\cos ^{-1} x\right)\left(\sin ^{-1} y\right)^2=\frac{\pi^4}{16}$

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad a+b^2=\frac{n \pi^2}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } a \cdot b^2=\frac{\pi^4}{16} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \begin{aligned} \therefore a-b & =\sqrt{\left(a+b^2\right)^2-4 a b} \\ & =\sqrt{\frac{n^2 \pi^4}{16}-\frac{4 \pi^4}{16}} \end{aligned} \end{aligned}\\ &\text { Now, least value of } n \text { can be } 2\\ &\begin{aligned} a-b^2 & =\sqrt{\frac{4 \pi^4}{16}-\frac{4 \pi^4}{16}} \\ \Rightarrow \quad a-b^2 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} \end{aligned} $

Adding Eqs. (i) and (iii), we get

$ \begin{aligned} & \quad 2 a=\frac{n \pi^2}{4} \quad \text { or } \quad 2 a=\frac{2 \pi^2}{4}(\because n=2) \\ & \text { or } \quad a=\frac{\pi^2}{4} \Rightarrow \cos ^{-1} x=\frac{\pi^2}{4} \\ & \Rightarrow \quad x=\cos \left(\frac{\pi^2}{4}\right) \end{aligned} $

Now, from Eq. (iii), $a=b$

$ \begin{aligned} & \left(\sin ^{-1} y\right)^2=\frac{\pi^2}{4} \text { or } \sin ^{-1} y= \pm \frac{\pi}{2} \\or\,\,\,\,\,\,\, & y=\sin \left( \pm \frac{\pi}{2}\right) \text { or } y= \pm 1 \end{aligned} $

Therefore, ordered pair is $\left(\cos \frac{\pi^2}{4}, \pm 1\right)$.

Given that,

$ x=\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right) $

As we know,

$ \begin{aligned} \tan ^{-1} x & +\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ \therefore \quad x & =\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{8+5}{40}}{\frac{40-1}{40}}\right) \\ x & =\tan ^{-1} \frac{13}{39}=\tan ^{-1}\left(\frac{1}{3}\right) \end{aligned} $

$ \therefore \tan x=\frac{1}{3} $

And $\sin x=\frac{1}{\sqrt{10}}$ and $\cos x=\frac{3}{\sqrt{10}}$

Putting values of $\sin x, \cos x, \tan x$

$ \frac{\sin x+\cos x}{\tan x}=\frac{\frac{1}{\sqrt{10}}+\frac{3}{\sqrt{10}}}{\frac{1}{3}}=\frac{\frac{4}{\sqrt{10}}}{\frac{1}{3}}=\frac{12}{\sqrt{10}} $

Given that, $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x=\log _e(f(x))$ for $|x|>1$

$ \begin{align*} & \because \quad \tanh ^{-1}\left(\frac{1}{x}\right)=\frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right) \\ & \left(\because \tanh ^{-1} x=\frac{1}{2} \log _e\left(\frac{1+x}{1-x}\right)\right) \\ & \tanh ^{-1}\left(\frac{1}{x}\right)=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \quad \operatorname{cot} \mathrm{h}^{-1} x=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

Adding the both Eqs. (i) and (ii), we get

$ \begin{aligned} \tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth} \mathrm{h}^{-1} x & =\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \\ & +\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \\ & =\log _e\left(\frac{x+1}{x-1}\right) \\ \tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x & =\log _e(f(x)) \end{aligned} $

On comparing, we get

$ f(x)=\frac{x+1}{x-1} $

At $x=-5, f(-5)=\frac{-5+1}{-5-1}=\frac{-4}{-6}$

$ \Rightarrow f(-5)=2 / 3 $

We have,

$ f(x)=\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right] $

$f(x)$ is defined of

$ \begin{array}{ll} & \log _5\left(x^2+7 x+15\right) \in[-1,1] \\ & -1 \leq \log _5\left(x^2+7 x+15\right) \leq 1 \\ & \frac{1}{5} \leq x^2+7 x+15 \leq 5 \\ \because & x^2+7 x+15 \leq 5 \\ \Rightarrow \quad & x^2+7 x+10 \leq 0 \\ & (x+5)(x+2) \leq 0 \\ \Rightarrow \quad & x \in[-5,-2] \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) & =\tan ^{-1} \alpha \\ \sum_{n=1}^k \tan ^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right) & =\tan ^{-1} \alpha \\ \Rightarrow \sum_{n=1}^k\left(\tan ^{-1}(n+2)-\tan ^{-1}(n+1)\right) & =\tan ^{-1} \alpha \\ \Rightarrow\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\left(\tan ^{-1} 4\right. & \left.-\tan ^{-1} 2\right) \\ \ldots . . \tan ^{-1}(k+2)-\tan ^{-1}(k+1) & =\tan ^{-1} \alpha \\ \tan ^{-1}(k+2)-\tan ^{-1} 2 & =\tan ^{-1} \alpha \end{aligned} \end{aligned} $

$ \begin{aligned} & \tan ^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right)=\tan ^{-1} \alpha \\ \Rightarrow & \frac{k}{1+2 k+4}=\alpha \Rightarrow \alpha=\frac{k}{2 k+5} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right) =\tan ^{-1}\left(\frac{2}{3}\right) \\ & \tan ^{-1} \frac{\left(\frac{x}{x-2}-\frac{x}{2 x-1}\right)}{1+\left(\frac{x}{x-2}\right)\left(\frac{x}{2 x-1}\right)}=\tan ^{-1} \frac{2}{3} \\ \Rightarrow \quad & \frac{2 x^2-x-x^2+2 x}{2 x^2-4 x-x+2+x^2}=\frac{2}{3} \\ \Rightarrow \quad & \frac{x^2+x}{3 x^2-5 x+2}=\frac{2}{3} \\ \Rightarrow \quad & \quad 3 x^2+3 x=6 x^2-10 x+4 \\ \Rightarrow \quad & \quad 3 x^2-13 x+4=0 \\ \Rightarrow \quad & \quad(3 x-1)(x-4)=0 \\ \Rightarrow \quad & x=\frac{1}{3}, 4, \quad x \in\left\{\frac{1}{3}, 4\right\} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2} \\ & \Rightarrow \\ & \sin ^{-1}\left(\frac{12}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{5}{x}\right) \\ & \Rightarrow \quad \sin ^{-1}\left(\frac{12}{x}\right)=\cos ^{-1}\left(\frac{5}{x}\right) \\ & \Rightarrow \quad \cos ^{-1}\left(\frac{\sqrt{x^2-144}}{x}\right)=\cos ^{-1}\left(\frac{5}{x}\right) \\ & \Rightarrow \quad \frac{\sqrt{x^2-144}}{x}=\frac{5}{x} \\ & \Rightarrow \quad \sqrt{x^2-144}=5 \Rightarrow x^2-144=25 \\ & \Rightarrow \quad x^2=144+25 \\ & \Rightarrow \quad x^2=169 \\ & \Rightarrow \quad x=13 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \begin{array}{c} \operatorname{cosec}^{-1}\left[\begin{array}{l} \left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}\right. \left.\left.\quad+\cos \frac{\alpha}{2}\right) \cot 5 \alpha \sec \frac{11 \alpha}{2}\right] \end{array}\right. \\ =\operatorname{cosec}^{-1}\left[\left(-\cos 2\left(\frac{\alpha-\pi}{4}\right)\right.\right. \left.\left.\quad+\cos \frac{\alpha}{2} \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right] \end{array} \\ & =\operatorname{cosec}{ }^{-1}\left[\left(-\sin \frac{\alpha}{2}\right.\right. \left.\left.\quad+\cos \frac{\alpha}{2} \frac{\cos 5 \alpha}{\sin 5 \alpha}\right) \sec \frac{11 \alpha}{2}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\operatorname{cosec}^{-1}\left[\left(\frac{-\sin 5 \alpha \sin \frac{\alpha}{2}+\cos 5 \alpha \cos \frac{\alpha}{2}}{\sin 5 \alpha}\right)\right. \left.\sec \frac{11 \alpha}{2}\right] \\ & =\operatorname{cosec}^{-1}\left[\frac{\cos \left(5 \alpha+\frac{\alpha}{2}\right)}{\sin 5 \alpha} \cdot \sec \frac{11 \alpha}{2}\right] \\ & =\operatorname{cosec}^{-1}\left[\operatorname{cosec} 5 \alpha \cdot \cos \frac{11 \alpha}{2} \sec \frac{11 \alpha}{2}\right] \\ & =\operatorname{cosec}^{-1}(\operatorname{cosec} 5 \alpha)=5 \alpha \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8} \\ \Rightarrow & \left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\frac{1}{2} \sec ^{-1} x=\frac{\pi}{8} \\ \Rightarrow & \tan ^{-1}\left[\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \times \frac{1}{8}}\right]+\frac{1}{2} \sec ^{-1} x=\frac{\pi}{8} \\ \Rightarrow & \tan ^{-1} \frac{13}{39}+\frac{1}{2} \sec ^{-1} x=\frac{\pi}{8} \\ \Rightarrow & 2 \tan ^{-1} \frac{13}{39}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & 2 \tan ^{-1} \frac{1}{3}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{3}{4}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{3}{4}+\tan ^{-1} \sqrt{x^2-1}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1}\left[\frac{\frac{3}{4}+\sqrt{x^2-1}}{1-\frac{3}{4} \sqrt{x^2-1}}\right]=\frac{\pi}{4} \end{aligned} \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \frac{3+4 \sqrt{x^2-1}}{4-3 \sqrt{x^2-1}}=1 \\ \Rightarrow & 3+4 \sqrt{x^2-1}=4-3 \sqrt{x^2-1} \\ \Rightarrow & 7 \sqrt{x^2-1}=1 \\ \Rightarrow & \sqrt{x^2-1}=\frac{1}{7} \\ \Rightarrow & x^2-1=\frac{1}{49} \\ \Rightarrow & x^2=\frac{50}{49} \end{array} $

Reason (R) $e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2-2 p-x=0$

We have, $x p^2-2 p-x=0$

$ \begin{aligned} \Rightarrow \quad p & =\frac{2 \pm \sqrt{4+4 x^2}}{2 x}=\frac{1 \pm \sqrt{1+x^2}}{x} \\ & =\frac{1+\sqrt{1+x^2}}{x}, \frac{1-\sqrt{1+x^2}}{x} \\ & =e^{\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)}, e^{\ln \left(\frac{1-\sqrt{1+x^2}}{x}\right)} \end{aligned} $

Since, $\operatorname{cosech}^{-1} x=\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)$

$\therefore e^{\operatorname{cosech}^{-1} x}$ is a solution of given equation.

$ \text { } \begin{aligned} Now,\,\, \operatorname{cosech}^{-1}(3) & =\ln \left(\frac{1+\sqrt{1+3^2}}{3}\right) \\ & =\ln \left(\frac{1+\sqrt{10}}{3}\right) \end{aligned} $

Let, ${\cot ^{ - 1}}(1 + x) = \alpha $

$ \Rightarrow 1 + x = \cot \alpha $

Now, let ${\tan ^{ - 1}}(x) = \beta $

$ \Rightarrow x = \tan \beta $

Given, $\sin ({\cot ^{ - 1}}(x)) = \cos ({\tan ^{ - 1}}(x))$

$ \Rightarrow \sin \alpha = \cos \beta $

From $\Delta$ABC, $\sin \alpha = {1 \over {\sqrt {1 + {x^2} + 2x + 1} }}$

$ = {1 \over {\sqrt {{x^2} + 2x + 2} }}$

From $\Delta$MNO, $\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}$

$\therefore$ ${1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}$

$ \Rightarrow {x^2} + 2x + 2 = {x^2} + 1$

$ \Rightarrow 2x + 2 = 1$

$ \Rightarrow 2x = - 1$

$ \Rightarrow x = -{1 \over 2}$

Here, $\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$

and $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$ as ${6 \over {11}} > {1 \over 2}$

$ \Rightarrow {\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 6}$

$\therefore$ $\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\pi \over 2} \Rightarrow \cos \alpha < 0$

Now, $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$

As ${4 \over 9} < {1 \over 2} \Rightarrow {\cos ^{ - 1}}\left( {{4 \over 9}} \right) > {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$

$\therefore$ $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right) > \pi $

$\therefore$ $\cos \beta < 0$ and $\sin \beta < 0$

Now, $\alpha + \beta $ is slightly greater than ${{3\pi } \over 2}$.

$\therefore$ $\cos (\alpha + \beta ) > 0$

$\begin{aligned} & \text { (P) Given } y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1] \\ & \Rightarrow \quad y(x)=4 \cos ^3\left(\cos ^{-1} x\right)-3 \cos \left(\cos ^{-1} x\right) \\ & \Rightarrow \quad y(x)=4 x^3-3 x \quad \text{... (i)}\\ & \Rightarrow \quad \frac{d y(x)}{d x}=12 x^2-3 \quad \text{... (ii)}\\ & \Rightarrow \quad \frac{d^2 y(x)}{d x^2}=24 x \quad \text{... (iii)}\\ \end{aligned}$

From equation (i), (ii) and (iii)

$\begin{aligned} \Rightarrow & \frac{1}{y(x)}\left[\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right] \\ & =\frac{1}{4 x^3-3 x}\left[\left(x^2-1\right) \cdot 24 x+x \cdot\left(12 x^2-3\right)\right] \\ & =\frac{1}{4 x^3-3 x}\left[24 x^3-24 x+12 x^3-3 x\right] \\ & =\frac{1}{4 x^3-3 x}\left[36 x^3-27 x\right] \end{aligned}$

$\begin{aligned} & =\frac{9\left(4 x^3-3 x\right)}{4 x^3-3 x} \\ & =9 \end{aligned}$

Hence P match with 4

(Q) Given $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \ldots . \mathrm{A}_n(n>2)$ be the vertices of a regular polygon of $n$ sides with centre at origin and $\vec{a}_k$ be the position vector of the point $A_k$

$\begin{aligned} & \text { Here, } \begin{aligned} &\left|\vec{a}_1\right|=\left|\vec{a}_2\right|=\left|\vec{a}_3\right|=\ldots=\left|\vec{a}_k\right| \\ &=\left|\vec{a}_{k+1}\right|=\cdots=\left|\vec{a}_n\right|=\lambda \text { (let) } \\ & \text { And } \angle \mathrm{A}_1 \mathrm{OA}_2=\angle \mathrm{A}_2 \mathrm{OA}_3=\cdots=\angle \mathrm{A}_k \mathrm{OA}_{k+1} \\ &=\cdots=\angle \mathrm{A}_n \mathrm{OA}_1=\frac{2 \pi}{n} \\ & \Rightarrow \vec{a}_k \times \vec{a}_{k+1}=\lambda \cdot \lambda \cdot \sin \frac{2 \pi}{n} \cdot \hat{p} \text { and } \\ & \vec{a}_k \cdot \vec{a}_{k+1}=\lambda \cdot \lambda \cdot \cos \frac{2 \pi}{n} \end{aligned} \end{aligned}$

Apply $\left|\sum_\limits{k=1}^{n-1} \vec{a}_k \times \vec{a}_{k+1}\right|=\left|\sum_\limits{k=1}^{n-1} \vec{a}_k \cdot \vec{a}_{k+1}\right|$

$\begin{aligned} & \left|\sum_{k=1}^{n-1} \lambda^2 \sin \frac{2 \pi}{n} \hat{p}\right|=\left|\sum_{k=1}^{n-1} \lambda^2 \cos \frac{2 \pi}{n}\right| \\ & \Rightarrow(n-1) \lambda^2\left|\sin \frac{2 \pi}{n}\right|=(n-1) \lambda^2\left|\cos \frac{2 \pi}{n}\right| \\ & \Rightarrow\left|\tan \frac{2 \pi}{n}\right|=1 \\ & \Rightarrow n_{\text {min }}=8 \\ \end{aligned}$

Hence $Q$ match with 3 .

(R) We know the equation of normal of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left(x_1, y_1\right)$ is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$.

$\therefore$ Equation of Normal of ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ at $p(h, 1)$ is $\frac{6 x}{h}-\frac{3 y}{1}=1$

Given, the normal at $p(h, 1)$ is perpendicular to the line $x+y=3$

$\Rightarrow \quad \frac{{\frac{6}{h}}}{3} \times(-1)=-1$

$\Rightarrow \quad \frac{2}{h}=1$

$\Rightarrow \quad h=2$

Hence, R match with 2.

$\begin{aligned} & \text { (S) } \tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1} \frac{2}{x^2} \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{\frac{1}{2 x+1}+\frac{1}{4 x+1}}{1-\frac{1}{2 x+1} \times \frac{1}{4 x+1}}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{6 x+2}{8 x^2+6 x+1-1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{3 x+1}{4 x^2+3 x}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right) \end{aligned}$

$\begin{aligned} & \Rightarrow & 3 x^3+x^2 & =8 x^2+6 x \\ & \Rightarrow & x\left[3 x^2-7 x-6\right] & =0 \\ & \Rightarrow & x\left[3 x^2-9 x+2 x-6\right] & =0 \\ & \Rightarrow & x(3 x+2)(x-3) & =0 \\ & \Rightarrow & x & =0, \frac{-2}{3}, 3 \end{aligned}$

So, $x=3$ is the only positive solution of given equation.

Hence, S match with 1.

Hint:

(i) Recall $\cos \left(\cos ^{-1} x\right)=x \forall x \in[-1,1]$ and $\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$

(ii) If angle between $\vec{a}$ and $\vec{b}$ is $\theta$, then $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$ and $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}$

(iii) The equation of normal of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left(x_1, y_1\right)$ is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$.

(iv) Recall $\tan ^{-1} \mathrm{~A}+\tan ^{-1} \mathrm{~B}=\tan ^{-1}\left(\frac{\mathrm{A}+\mathrm{B}}{1-\mathrm{AB}}\right)$ only when A.B < 1

For P

$\text { Let } t=\sqrt{\frac{1}{y^2}\left(\frac{\cos \left(\tan ^{-1} y\right)+y \cdot \sin \left(\tan ^{-1} y\right)}{\cot \left(\sin ^{-1} y\right)+\tan \left(\sin ^{-1} y\right)}\right)^2+y^4}$

We know that

$\begin{gathered} \tan ^{-1} y=\cos ^{-1}\left(\frac{1}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) \\ \text { and } \sin ^{-1} y=\cot ^{-1}\left(\frac{\sqrt{1-y^2}}{y}\right)=\tan ^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) \end{gathered}$

$\therefore \quad t = \sqrt {{1 \over {{y^2}}}{{\left[ {{{\cos \left( {{{\cos }^{ - 1}}{1 \over {\sqrt {1 + {y^2}} }}} \right) + y\,.\,\sin \left( {{{\sin }^{ - 1}}{y \over {\sqrt {1 + {y^2}} }}} \right)} \over {\cot \left( {{{\cot }^{ - 1}}{{\sqrt {1 - {y^2}} } \over y}} \right) + \tan \left( {{{\tan }^{ - 1}}{y \over {\sqrt {1 - {y^2}} }}} \right)}}} \right]}^2} + {y^4}} $

$\begin{aligned} & \Rightarrow t=\sqrt{\frac{1}{y^2} \times\left[\frac{\frac{1}{\sqrt{1+y^2}}+\frac{y^2}{\sqrt{1+y^2}}}{\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}}\right]^2+y^4} \\ & \Rightarrow t=\sqrt{\frac{1}{y^2} \times\left[\frac{\sqrt{1+y^2}}{\frac{1-y^2+y^2}{y \sqrt{1-y^2}}}\right]^2+y^4} \\ & \Rightarrow t=\sqrt{\frac{1}{y^2} \times y^2\left(1+y^2\right)\left(1-y^2\right)+y^4} \\ & \Rightarrow t=\sqrt{\left(1-y^4\right)+y^4}=1 \\ \end{aligned}$

Hence, P match with 4.

For Q.

Given $\cos x+\cos y+\cos z=0$

$\Rightarrow \cos x+\cos y=-\cos z$

On squaring both side

$\Rightarrow \cos ^2 x+\cos ^2 y+2 \cos x \cdot \cos y=\cos ^2 z\quad \text{... (i)}$

Also given $\sin x+\sin y+\sin z=0$

$\Rightarrow \sin x+\sin y=-\sin z$

On squaring both side

$\Rightarrow \sin ^2 x+\sin ^2 y+2 \sin x \cdot \sin y=\sin ^2 z \quad \text{... (ii)}$

Add equation (i) and (ii)

$\begin{gathered} \Rightarrow\left(\cos ^2 x+\sin ^2 x\right)+\left(\cos ^2 y+\sin ^2 y\right) \\ +2[\cos x \cdot \cos y+\sin x \cdot \sin y] \\ =\cos ^2 z+\sin ^2 z \end{gathered}$

$\begin{array}{lr} \Rightarrow & 1+1+2 \cos (x-y)=1 \\ \Rightarrow & \cos (x-y)=\frac{-1}{2} \\ \Rightarrow & 2 \cos ^2\left(\frac{x-y}{2}\right)-1=\frac{-1}{2} \end{array}$

$\begin{array}{ll} \Rightarrow & \cos ^2\left(\frac{x-y}{2}\right)=\frac{1}{4} \\ \Rightarrow & \cos \left(\frac{x-y}{2}\right)= \pm \frac{1}{2} \end{array}$

Hence, Q match with 3.

For R

Given,

$\begin{aligned} & \cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \cdot \sin 2 x \cdot \sec x \\ & =\cos x \cdot \sin 2 x \cdot \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x \end{aligned}$

$\begin{aligned} \Rightarrow & {\left[\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{\pi}{4}+x\right)\right] } \\ & \cos 2 x+(\sin x-\cos x) \sin 2 x \cdot \sec x=0 \end{aligned}$

$\begin{aligned} \Rightarrow & 2 \cdot \sin \frac{\pi}{4} \cdot \sin x \cdot \cos 2 x+(\sin x-\cos x) \\ & \frac{2 \sin x \cdot \cos x}{\cos x}=0 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \sqrt{2} \cos 2 x=2(\cos x-\sin x) \\ & \Rightarrow \quad\left(\cos ^2 x-\sin ^2 x\right)=\sqrt{2} (\cos x-\sin x) \\ \end{aligned}$

$\begin{array}{rlrl} \Rightarrow & \cos x+\sin x =\sqrt{2} \\ \Rightarrow & (\cos x+\sin x)^2 =(\sqrt{2})^2 \\ \Rightarrow & \cos ^2 x+\sin ^2 x+2 \sin x \cdot \cos x =2 \\ \Rightarrow & 1+\sin 2 x =2 \\ \Rightarrow & \sin 2 x =1 \\ \Rightarrow & 2 x =\frac{\pi}{2} \\ \Rightarrow & x =\frac{\pi}{4} \\ \Rightarrow & \sec x =\sec \frac{\pi}{4}=\sqrt{2} \end{array}$

Hence, R match with 2

For S

Given,

$\begin{aligned} \cot \left(\sin ^{-1} \sqrt{1-x^2}\right) & =\sin \left(\tan ^{-1}(x \sqrt{6})\right), x \neq 0 \\ \Rightarrow \cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\right) & =\sin \left(\sin ^{-1} \frac{x \sqrt{6}}{\sqrt{1+6 x^2}}\right) \\ \Rightarrow \quad \frac{x}{\sqrt{1-x^2}} & =\frac{x \sqrt{6}}{\sqrt{1+6 x^2}} \end{aligned}$

$ \begin{array}{rlrl} \Rightarrow & \frac{1}{\sqrt{1-x^2}} & =\frac{\sqrt{6}}{\sqrt{1+6 x^2}} \\ \Rightarrow & \text { on squaring both sides } \\ \Rightarrow & \frac{1}{1-x^2} =\frac{6}{1+6 x^2} \\ \Rightarrow & 1+6 x^2=6-6 x^2 \\ \Rightarrow & x^2 =\frac{5}{12} \\ \Rightarrow & x= \pm \sqrt{\frac{5}{12}}= \pm \frac{1}{2} \sqrt{\frac{5}{3}} \end{array}$

Hence, S match with 1.

Hints :

(i) Recall the method of conversion of one inverse trigonometric function into other inverse trigonometric function as

$\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}$

(ii) Recall $\cos \left(\cos ^{-1} x\right)=x, \sin \left(\sin ^{-1} x\right)=x\tan \left(\tan ^{-1} x\right)=x \text { etc }$

(iii) Recall the following trigonometric identities

(A) $\sin ^2 x+\cos ^2 x=1$

(B) $\sin 2 x=2 \sin x \cdot \cos x$

(C) $\cos 2 x=\cos ^2 x-\sin ^2 x=2 \cos ^2 x-1= 1-2 \sin ^2 x$

(D) $\cos (A-B)-\cos (A+B)=2 \sin A \cdot \sin B$

$\cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + 2 \times {{n(n + 1)} \over 2}} \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}(1 + n(n + 1))} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {{{n(n + 1) + 1} \over {(n + 1) - n}}} \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}n - {{\cot }^{ - 1}}(n + 1)} } \right)$

$ = \cot (({\cot ^{ - 1}}1 + {\cot ^{ - 1}}2 + {\cot ^{ - 1}}3 + .... + {\cot ^{ - 1}}23) - ({\cot ^{ - 1}}2 + {\cot ^{ - 1}}3 + .... + {\cot ^{ - 1}}23 + {\cot ^{ - 1}}24))$

$ = \cot ({\cot ^{ - 1}}1 - {\cot ^{ - 1}}24)$

$ = \cot \left( {{{\cot }^{ - 1}}{{24 \times 1 + 1} \over {24 - 1}}} \right) = {{25} \over {23}}$

Given that,

$0 < x < 1$,

$\sqrt {1 + {x^2}} {[{\{ x\cos ({\cot ^{ - 1}}x) + \sin ({\cot ^{ - 1}}x)\} ^2} - 1]^{{1 \over 2}}}$

Find : value of this expression

Here, $0 < x < 1$,

${\cot ^{ - 1}}x = {\sin ^{ - 1}}\left( {{1 \over {\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {{x \over {\sqrt {1 + {x^2}} }}} \right)$

Now, $\sqrt {1 + {x^2}} {[{\{ x\cos ({\cot ^{ - 1}}x) + \sin ({\cot ^{ - 1}}x)\} ^2} - 1]^{{1 \over 2}}}$

$\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cos }^{ - 1}}\left( {{x \over {1 + {x^2}}}} \right)} \right) + \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt {1 + {x^2}} }}} \right)} \right\}}^2}} \right]^{{1 \over 2}}}$

$ = \sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\,.\,{x \over {\sqrt {1 + {x^2}} }} + {1 \over {\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right]^{{1 \over 2}}}$

$ = \sqrt {1 + {x^2}} {\left[ {{{\left\{ {{{{x^2} + 1} \over {\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right]^{{1 \over 2}}}$

$ = \sqrt {1 + {x^2}} {[{x^2} + 1 - 1]^{{1 \over 2}}}$

$ = x\sqrt {1 + {x^2}} $

which is the value of the expression.