Question

Solve:
${\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)$

Answer 1

$\begin{array}{c}{\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)\\ letx=\tan \theta ,\theta ={\tan }^{-1}x\\ {\tan }^{-1}\left(\frac{5x-x}{1+5x\cdot x}\right)={\tan }^{-1}\left(\frac{5\tan \theta -\tan \theta }{1+5{\tan }^2\theta }\right)\\ {\tan }^{-1}\tan \left(4\theta \right)\left[\tan \left(A-B\right)=\left(\frac{\tan A-\tan B}{1+\tan A\cdot \tan B}\right)\right]\\ \left[{\tan }^{-1}\tan \theta =\theta \right]\\ =4\theta =4{\tan }^{-1}xAns.\end{array}$