Trigonometric-Identities Question 380
Question: $ \tan 9{}^\circ -\tan 27{}^\circ -\tan 63{}^\circ +\tan 81{}^\circ = $
[Roorkee 1989]
Options:
A) 1/2
B) 2
C) 4
D) 8
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tan 9^{o}-\tan 27^{o}-\tan 63^{o}+\tan 81^{o} $ $ =\tan 9^{o}-\tan 27^{o}-\cot 27^{o}+\cot 9^{o} $ $ =(\tan 9^{o}+\cot 9^{o})-(\tan 27^{o}+\cot 27^{o}) $ $ =\frac{\cos (9^{o}-9^{o})}{\sin 9^{o}\cos 9^{o}}-\frac{\cos (27^{o}-27^{o})}{\sin 27^{o}.\cos 27^{o}}=\frac{2}{\sin 18^{o}}-\frac{2}{\sin 54^{o}} $ $ =2,{ \frac{\sin 54^{o}-\sin 18^{o}}{\sin 18^{o},\sin 54^{o}} }=2.\frac{2,.,\cos 36^{o}.,\sin 18^{o}}{\sin 18^{o}.,\sin 54^{o}}=4 $
BETA
