Amit M Agarwal Solutions for Chapter: Inverse Trigonometric Functions, Exercise 3: Objective Questions

If in $\Delta PQR, \text{angles} P={\sin }^{-1}\left(\alpha \right), Q={\sin }^{-1}\left(\beta \right)$ and $R={\sin }^{-1}\left(\gamma \right)$, then  $\alpha \sqrt{1-{\beta }^2}\sqrt{1-{\gamma }^2}+\beta \sqrt{1-{\alpha }^2}\sqrt{1-{\gamma }^2}+\gamma \sqrt{1-{\alpha }^2}\sqrt{1-{\beta }^2}$ is:

If ${\sin }^{-1}\left(\sin x\right)=\pi -x,$ then $x$ belongs to:

The number of real solutions of ${\cos }^{-1}a+{\cos }^{-1}2a=-\pi$ is:

If $a,b$ are the roots of the equation; 

$6x^2+11x+3=0$

${\tan }^{-1}\alpha +{\tan }^{-1}\beta ,$ where $\alpha >0, \beta >0, \alpha \beta >1$ is equal to

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

$\tan \left(\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\alpha \right)+\tan \left(\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\alpha \right), x\ne 0,$ is equal to :

The number of solutions of the equation;
$1+a^2+2a\sin \left({\cos }^{-1}b\right)=0$ is: