Question

If $si{n}^{-1}\frac{2a}{1+a^2}-co{s}^{-1}\frac{1-b^2}{1+b^2}={tan}^{-1}\frac{2x}{1-x^2}$, then prove that $x=\frac{a-b}{1+ab}$

Answer 1

Let $a=\tan {\theta }_1,b=\tan {\theta }_2$ and $x=\tan \theta$

${\sin }^{-1}\left(\frac{2a}{1+a^2}\right)-{\cos }^{-1}\left(\frac{1-b^2}{1+b^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

$\Rightarrow {\sin }^{-1}\left(\frac{2\tan {\theta }_1}{1+{\tan }^2}{\theta }_1\right)-{\cos }^{-1}\left(\frac{1-{\tan }^2{\theta }_2}{1+{\tan }^2{\theta }_2}\right)={\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)$

$\Rightarrow {\sin }^{-1}\left(\sin 2{\theta }_1\right)-{\cos }^{-1}\left(\cos 2{\theta }_2\right)={\tan }^{-1}\left(\tan 2\theta \right)$

$\Rightarrow 2{\theta }_1-2{\theta }_2=2\theta$

$\Rightarrow {\theta }_1-{\theta }_2=\theta$

$\Rightarrow \tan ({\theta }_1-{\theta }_2)=\tan \theta$

$\Rightarrow \frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}=\tan \theta$

$\Rightarrow \frac{a-b}{1+ab}=\tan \theta$

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

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

Hence proved.