Trigonometric Identities Question 114
Question: $ \tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha = $
[IIT 1988; MP PET 1991]
Options:
A) $ \tan \alpha $
B) $ \tan 2\alpha $
C) $ \cot \alpha $
D) $ \cot 2\alpha $
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Answer:
Correct Answer: C
Solution:
$ \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 4a(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 $ .
BETA
