Question

Prove $\cot \alpha -\tan \alpha -2\tan 2\alpha -4\tan 4\alpha =8\cot 8\alpha .$

Answer 1

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

$⟹\cot \alpha -\tan \alpha -2\tan 2\alpha =2\times \frac{\cos 2\alpha }{\sin 2\alpha }-\frac{2\sin 2\alpha }{\cos 2\alpha },$

$⟹\cot \alpha -\tan \alpha -2\tan 2\alpha =2\times \frac{\cos 4\alpha }{\sin 2\alpha \times \cos 2\alpha }=\frac{4\cos 4\alpha }{\sin 4\alpha }$

$⟹\cot \alpha -\tan \alpha -2\tan 2\alpha -4\tan 4\alpha =\frac{4\cos 4\alpha }{\sin 4}-\frac{4\sin 4\alpha }{\cos 4\alpha }$

$⟹\cot \alpha -\tan \alpha -2\tan 2\alpha -4\tan 4\alpha =\frac{4\cos 8\alpha }{\sin 4\cos 4\alpha }=\frac{8\cos 8\alpha }{\sin 8\alpha }$

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