Question

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

• A. $\tan \alpha$
• B. $\tan 2\alpha$
• C. $\cot \alpha$
• D. $\cot 2\alpha$

Answer 1

The correct option is C $\cot \alpha$

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha$
$=\tan \alpha +2\tan 2\alpha +4[\frac{\sin 4\alpha }{\cos 4\alpha }+2\frac{\cos 8\alpha }{\sin 8\alpha }]$

$=\tan \alpha +2\tan 2\alpha +4[\frac{\cos 4\alpha \cos 8\alpha +\sin 4\alpha \sin 8\alpha +\cos 4\alpha \cos 8\alpha }{\sin 8\alpha \cos 4\alpha }]$

$=\tan \alpha +2\tan 2\alpha +4[\frac{\cos 4\alpha +\cos 4\alpha \cos 8\alpha }{\sin 8\alpha \cos 4\alpha }]$

$=\tan \alpha +2\tan 2\alpha +4[\frac{\cos 4\alpha (1+\cos 8\alpha )}{\cos 4\alpha \sin 8\alpha }]$

$=\tan \alpha +2\tan 2\alpha +4[\frac{2{\cos }^{2}4\alpha }{2\sin 4\alpha \cos 4\alpha }]$

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

$=\tan \alpha +2(\tan 2\alpha +2\cot 4\alpha )$

$=\tan \alpha +2[\frac{\sin 2\alpha }{\cos 2\alpha }+2\frac{\cos 4\alpha }{\sin 4\alpha }]$

$=\tan \alpha +2[\frac{\cos 2\alpha (1+\cos 4\alpha )}{\sin 4\alpha \cos 2\alpha }]$

$=\tan \alpha +2\cot 2\alpha =\frac{\sin \alpha }{\cos \alpha }+\frac{2\cos 2\alpha }{\sin 2\alpha }$

$=\frac{\cos \alpha +\cos \alpha \cos 2\alpha }{\sin 2\alpha \cos \alpha }$

$=\frac{1+\cos 2\alpha }{\sin 2\alpha }=\frac{2{\cos }^2\alpha }{2\sin \alpha \cos \alpha }=\cot \alpha$