Question

Prove that: $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$

Answer 1

$\begin{array}{c}Let{\tan }^{-1}x=\theta \\ \Rightarrow \tan \theta =x\\ weknowthat\\ {\tan }^2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\\ \Rightarrow {\tan }^2\theta =\frac{2x}{1-x^2}\\ \Rightarrow 2\theta ={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\\ \Rightarrow 2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\end{array}$

Hence, proved.