Basics of Inverse Trigonometric Functions

Suppose $f\left(x\right)=\tan \left({\sin }^{-1}\left(2x\right)\right)$. Find range of $f\left(x\right)$.

Evaluate:   ${\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^1\left(\frac{1}{5}\right)+{\tan }^1\left(\frac{1}{8}\right)$

Evaluate :  ${\tan }^{-1}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]$

Find the value of ${\sin }^{-1}\left(\sin \frac{4\pi }{5}\right).$

The principal value of ${\tan }^{-1}\sqrt{3}-{\sec }^{-1}\left(-2\right)$ would be

Which of the following is the value of the given expression : ${\tan }^{-1}\left(1\right)+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(-\frac{1}{2}\right)$

The value of $x$ for which  $\sin \left[{\cot }^{-1}\left(1+x\right)\right]=\cos \left({\tan }^{-1}x\right)$

The principal value of  ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$ is

${\tan }^{-1}\left[\frac{1}{\sqrt{3}}\sin \frac{5\pi }{2}\right]{\sin }^{-1}\left[\cos \left({\sin }^{-1}\frac{\sqrt{3}}{2}\right)\right]=$

$\cos \left[{\cot }^{-1}\left(-\sqrt{3}\right)+\frac{\pi }{6}\right]=$

If ${\sin }^{-1}\left(\tan \frac{\pi }{4}\right)-{\sin }^{-1}\sqrt{\frac{3}{x}}=\frac{\pi }{6}$, then $x$ is a root of the equation

Find the principal values of the following question:

${\text{sin}}^{\mathrm{-1 }}\left(-\frac{1}{2}\right)$

The values of $x$ for which the function $f\left(x\right)={\tan }^{-1}x-{\cot }^{-1}x+{\cos }^{-1}\left(2-x\right)$ is defined, is

If $f\left(x\right)={\sin }^{-1}\left(\mathrm{cosec}\left({\sin }^{-1}x\right)\right)+{\cos }^{-1}\left(\sec \left({\cos }^{-1}x\right)\right),$ then $f\left(x\right)$ takes:

Find the number of solutions of the equation ${\sin }^{-1}\left(\mathrm{sinx}\right)+{\cos }^{-1}\left(\mathrm{cosx}\right)=\mathrm{x},\forall \mathrm{x}\in (0,\mathrm{\pi }]$ .

If $x$ takes all permissible negative values, then ${\sin }^{-1}x$ is equal to

The solution set of the inequality $\left(\sin x+{\cos }^{-1}x\right)-\left(\cos x-{\sin }^{-1}x\right)\ge \frac{\pi }{2}$ is equal to

If $f\left(x\right)={\sin }^{-1}\left\{\frac{\sqrt{3}}{2}x-\frac{1}{2}\sqrt{1-x^2}\right\}, -\frac{1}{2}\le x\le 1$, then $f\left(x\right)$ is equal to