Question

Prove that -
$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha =\cot \alpha$

Answer 1

$\tan \theta -\cot \theta =\frac{{\sin }^2\theta -{\cos }^2\theta }{\sin \theta \cos \theta }$

$=\frac{-2\cos 2\theta }{\sin 2\theta }$

$\tan \theta -\cot \theta =-2\cot 2\theta$

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha$

$=\tan \alpha -\cot \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha +\cot \alpha$

$=-2\cot 2\alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha +\cot \alpha$

$=-4\cot 4\alpha +4\tan 4\alpha +8\cot 8\alpha +\cot \alpha$

$=-8\cot 8\alpha +8\cot 8\alpha +\cot \alpha$

$=\cot \alpha$