Question

If ${\tan }^3\theta =\tan \varphi$ and $\tan 2\theta =2\tan \alpha ,$ prove that -
$\theta +\varphi =n\pi +\alpha$.

Answer 1

$\tan (\theta +\varphi )=\frac{\tan \theta +tan\varphi }{1-tan\theta tan\varphi }$
$\frac{\tan \theta (1+{\tan }^2\theta )}{1-{\tan }^4\theta }=\frac{1}{2}\cdot \frac{2tan\theta }{1-{\tan }^2\theta }$
$\therefore \tan (\theta +\varphi )=\frac{1}{2}2\tan \theta =\frac{1}{2}2\tan \alpha =\tan \alpha$
$\therefore \theta +\varphi =n\pi +\alpha$