Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=\tan m \\ b=\tan n \\ x=\tan y \end{gathered}$$

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

$\therefore \frac{a-b}{1+ab}=x$