Identities of Inverse Trigonometric Functions

If $y={\tan }^{-1}\frac{1}{x^2+x+1}+{\tan }^{-1}\frac{1}{x^2+3x+3}+{\tan }^{-1}\frac{1}{x^2+5x+7}+{\tan }^{-1}\frac{1}{x^2+7x+13}+\dots \dots \dots \dots$ up to $n$ terms, then $\frac{dy}{dx}$ is

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

${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4},$ then the value of $x$ could be

$\cos \left({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2}\right)$  would be equal to:

The solution of the equation ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$ would be:

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

The value of  $\tan \left[{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]$ is

$\tan \left[\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)+\frac{1}{2}{\cos }^{-1}\left(\frac{15}{17}\right)\right]$ is equal to

$\cos \left({\cot }^{-1}\left(cosec\left({\cos }^{-1}a\right)\right)\right)=\_\_\_\_\_$

If $y=\sum _{r=1}^n{\tan }^{-1}\left(\frac{1}{x^2+(2r-1)x+r(r-1)+1}\right)$, then the value of $\frac{dy}{dx}$ at $x=0$ is 

$2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}=$

Solve the equation ${\cos }^{-1}x+{\cos }^{-1}2x=\frac{2\pi }{3}$.

Write the simplest form of ${\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), 0<x<\frac{\pi }{2}$.

Value of ${\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{12}{13}\right)$ is

$2 {\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$ is equal to 

${\sin }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)=$

$2 {\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$ is equal to