• # question_answer $\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot \,8\alpha =$ [IIT 1988; MP PET 1991] A) $\tan \alpha$ B) $\tan 2\alpha$ C) $\cot \,\alpha$ D) $\cot \,2\alpha$

$\tan \alpha +2\tan \,\,2\alpha +4\tan \,\,4\alpha +8\cot \,8\alpha$ $=\tan \alpha +2\tan \,2\alpha +4\left[ \frac{\sin 4\alpha }{\cos 4\alpha }+2\frac{\cos \,8\alpha }{\sin \,8\alpha } \right]$ $=\tan \alpha +2\tan 2\alpha +$            $4\left[ \frac{\cos \,4\alpha \,\cos \,8\alpha +\sin \,4\alpha \,\sin \,8\alpha +\cos \,4\alpha \cos \,8\alpha }{\sin \,8\alpha \,\cos \,4\alpha } \right]$ $=\tan \,\alpha +2\tan \,2\alpha +4\left[ \frac{\cos \,4\alpha +\cos \,4\alpha \,\cos \,8\alpha }{\sin \,8\alpha \cos \,4\alpha } \right]$ $=\tan \,\alpha +2\,\tan \,2\alpha +4\,\left[ \frac{\cos \,\,4a(1+\cos \,8\alpha )}{\cos \,4\alpha \sin \,8\alpha } \right]$ $=\tan \alpha +2\tan \,2\alpha +4\left[ \frac{2{{\cos }^{2}}4\alpha }{2\sin \,4\alpha \,\,\cos \,\,4\alpha } \right]$ $=\tan \,\alpha +2\tan \,2\alpha +4\cot \,4\alpha$ $=\tan \alpha +2(\tan 2\alpha +2\cot 4\alpha )$ $=\tan \,\alpha +2\left[ \frac{\sin \,\,2\alpha }{\cos 2\alpha }+2\frac{\cos \,4\alpha }{\sin \,4\alpha } \right]$ $=\tan \,\alpha +2\left[ \frac{\cos \,2\alpha (1+\cos \,4\alpha )}{\sin \,4\alpha \cos \,2\alpha } \right]$ $=\tan \alpha +2\cot 2\alpha =\frac{\sin \,\alpha }{\cos \,\alpha }+\frac{2\cos \,2\alpha }{\sin \,2\alpha }$ $=\frac{\cos \,\alpha +\cos \alpha \cos \,2\alpha }{\sin \,2\alpha \cos \alpha }$ $=\frac{1+\cos \,2\alpha }{\sin \,2\alpha }=\frac{2{{\cos }^{2}}\alpha }{2\sin \alpha \cos \alpha }=\cot \,\alpha$.