Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 2: Exercise-2

If $\mathrm{X}=\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\sec \left({\sin }^{-1}\mathrm{a}\right)\right)\right)\right)\right)$ and $Y=\sec \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}\left(\mathrm{cosec}\left({\cos }^{-1}a\right)\right)\right)\right)\right)$, where $0\le a<1$. Find the relation between $X$ and $Y$. Then

The expression $\frac{1}{\sqrt{2}}\left(\frac{\sin \left({\cot }^{-1}\left(\cos \left({\tan }^{-1}\mathrm{t}\right)\right)\right)}{\cos \left({\tan }^{-1}\left(\sin \left({\cot }^{-1}\sqrt{2}\mathrm{t}\right)\right)\right)}\right)\left(\sqrt{\frac{1+2{\mathrm{t}}^2}{2+{\mathrm{t}}^2}}\right)$ can take the value

If $0<x<1,$ then ${\tan }^{-1}\frac{\sqrt{1-x^2}}{1+x}$ is equal to

If $f\left(x\right)={\cos }^{-1}x+{\cos }^{-1}\left\{\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\right\},$ then

$\sum _{n=1}^{\infty }{\tan }^{-1}\frac{4n}{n^4-2n^2+2}$ is equal to

If ${\sin }^2\left(2{\cos }^{-1}\left(\tan x\right)\right)=1$ then $x$ may be

If ${\sin }^{-1}x+2{\cot }^{-1}\left(y^2-2y\right)=2\mathrm{\pi },$ then

$x\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right), {cosec}^{-1}\left(cosecx\right)$ is