NIMCET 2017 — Mathematics PYQ

Concept:
Double angle formula:
$\sin 2x=\frac{2\tan x}{1+{\tan }^2}$
Addition formula:
$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

Calculation:

$\Gamma$he given identity is ${\sin }^{-}1\left(\frac{2\mathrm{a}}{1+{\mathrm{a}}^2}\right)+{\sin }^{-1}\left(\frac{2\mathrm{b}}{1+{\mathrm{b}}^2}\right)=2{\tan }^{-1}$n.
Let a= $\tan y_1$ and b= $\tan y_2$. 
Therefore, the given equation becomes:
${\sin }^{-1}\left(\frac{2\mathrm{a}}{1+{\mathrm{a}}^2}\right)+{\sin }^{-1}\left(\frac{2\mathrm{b}}{1+{\mathrm{b}}^2}\right)=2{\tan }^{-1}$n
${\sin }^{-1}\left(\frac{2(\tan {\mathbf{y}}_1)}{1+(\tan {\mathbf{y}}_1{)}^2}\right)+{\sin }^{-1}\left(\frac{2(\tan {\mathbf{y}}_2)}{1+(\tan {\mathbf{y}}_2{)}^2}\right)=2{\tan }^{-1}\mathbf{m}$
${\sin }^{-1}(\sin 2{\mathbf{y}}_1)+{\sin }^{-1}(\sin 2{\mathbf{y}}_2)=2{\tan }^{-1}\mathbf{n}$
$2{\mathbf{y}}_1+2{\mathbf{y}}_2=2{\tan }^{-1}\mathbf{n}$
${\mathbf{y}}_1+{\mathbf{y}}_2={\tan }^{-1}\mathbf{n}$ $\tan ({\mathbf{y}}_1+{\mathbf{y}}_2)=\mathbf{n}$ $\frac{\tan {\mathrm{y}}_1+\tan {\mathrm{y}}_2}{1-\tan {\mathrm{y}}_1\tan {\mathrm{y}}_2}=$n $\frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}=\mathbf{n}$
Therefore, n$=\frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}.$