D.P. GUPTA, Preetima Bajpai and, Sanjeev Kumar Jha Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercise 1

The solutions of the equation  ${\tan }^{-1}\left(x+1\right)+{\tan }^{-1}\left(x-1\right)={\tan }^{-1}\frac{8}{31}$ are 

$2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+2{\tan }^{-1}\frac{1}{8}$ is equal to 

The equation ${\sin }^{-1}\frac{5}{x}+{\sin }^{-1}\frac{12}{x}=\frac{\pi }{2}$ has 

The solution of the inequality ${\log }_{\frac{1}{2}}{\sin }^{-1}x>{\log }_{\frac{1}{2}}{\cos }^{-1}x$ is 

If $x\in \left(\frac{11\pi }{2},\frac{15\pi }{2}\right)$, then ${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)=$

If $k\le {\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x\le K$, then

The set of values of $k$ for which $x^2-kx+{\sin }^{-1}\left(\sin 4\right)>0$ for all real $x$ is

If ${\cos }^{-1}x>{\sin }^{-1}x$, then the set of all values of $x$ is