Question

$\mathrm{If}{\sin }^{-1}\frac{2a}{1+a^2}+{\sin }^{-1}\frac{2b}{1+b^2}=2{\tan }^{-1}x,\mathrm{prove}\mathrm{that}x=\frac{a+b}{1-ab}.$

Answer 1

Let:
$$\begin{gathered} a=\tan z \\ b=\tan y \end{gathered}$$

Then,
$$\begin{gathered} {\sin }^{-1}\frac{2a}{1+a^2}+{\sin }^{-1}\frac{2b}{1+b^2}=2{\tan }^{-1}x \\ \Rightarrow {\sin }^{-1}\frac{2\tan z}{1+{\tan }^{2}z}+{\sin }^{-1}\frac{2\tan y}{1+{\tan }^{2}y}=2{\tan }^{-1}x \\ \Rightarrow {\sin }^{-1}\left(\sin 2z\right)+{\sin }^{-1}\left(\sin 2y\right)=2{\tan }^{-1}x\left[\because \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}\right] \\ \Rightarrow 2z+2y=2{\tan }^{-1}x \\ \Rightarrow {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}x\left[\because a=\tan z\mathrm{and}b=\tan y\right] \\ \Rightarrow {\tan }^{-1}\frac{a+b}{1-ab}={\tan }^{-1}x\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}\right] \\ \Rightarrow x=\frac{a+b}{1-ab} \end{gathered}$$