Properties of Inverse Trigonometric Functions

Evaluate: $\cot \left({\sin }^{-1}\frac{1}{\sqrt{5}}+{\sin }^{-1}\frac{2}{\sqrt{5}}\right)$

If $y={\cos }^{-1}x, -1⩽x⩽1,0⩽y⩽\mathrm{\pi }$ then prove that $\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}.$

If $a_1=1, a_2,a_3,a_4,.......$ be consecutive natural numbers. Then ${\tan }^{-1}\left(\frac{1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{1}{1+a_2{a}_3}\right)+.........+{\tan }^{-1}\left(\frac{1}{1+a_{2021}{a}_{2022}}\right)$ is equal to $?$

Find the value of ${\cos }^{-1}\left(\cos \frac{13\mathrm{\pi }}{6}\right)$.

If $A=\sin \left[{\cot }^{-1}\left\{\cos \left({\tan }^{-1}x\right)\right\}\right],$ then

Find:

${\sin }^{-1}\left(\frac{3}{5}\right)+ {\mathrm{cosec}}^{-1}\left(\frac{5}{4}\right)$

Simplify: $\tan \left(\frac{1}{2}{\cos }^{-1}\frac{2}{\sqrt{5}}\right)$

$x\in R, {\cot }^{-1}x=$

${\cot }^{-1}x={\cos }^{-1}(-1)-{\mathrm{cosec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Based on the above equation, find ${\tan }^{-1}\left(\frac{1}{x}\right)$ using the principal values of the inverse trigonometric functions. Show your work. 

${\mathrm{Sec}}^2\left({\tan }^{-1}2\right)+{\mathrm{cosec}}^2\left({\cot }^{-1}3\right)=?$

${\tan }^{-1}\left(\frac{1-\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)\tan \theta }{\tan \left(\frac{\mathrm{\pi }}{4}\right)+\tan \theta }\right)$

${\tan }^{-1}\sqrt{3}-{\sec }^{-1}\left(-2\right)$ is equal to

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

$\text{The equality }{\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}\text{, holds true for }\mathbf{x}\in \mathbf{R}$

Solve the equation:

$4{\sin }^{-1}x=\mathrm{\pi }-{\cos }^{-1}x$

For $x\in \left(0,\frac{\mathrm{\pi }}{2}\right)$, the expression ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\right)=$

If $2{\sin }^{-1}x+{\cos }^{-1}x=\frac{2\mathrm{\pi }}{3},$ then the value of $x$ is

The value of $\sum _{r=2}^{\infty }{\tan }^{-1}\left(\frac{1}{r^2-5r+7}\right),$ is

Value of $\sin \left[{\cot }^{-1}\left\{\tan \left({\cos }^{-1}x\right)\right\}\right],x\ne 0$ is 

Evaluate: $\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{n^2+n+1}\right).$