Trigonometric Identities Question 15
Question: $ \tan \frac{2\pi }{5}-\tan \frac{\pi }{15}-\sqrt{3}\tan \frac{2\pi }{5}\tan \frac{\pi }{15} $ is equal to
Options:
A) $ -\sqrt{3} $
B) $ \frac{1}{\sqrt{3}} $
C) 1
D) $ \sqrt{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ \frac{\tan \frac{6\pi }{15}-\tan \frac{\pi }{15}}{1+\tan \frac{6\pi }{15}\tan \frac{\pi }{15}}=\tan \frac{\pi }{3} $
$ \Rightarrow \tan \frac{6\pi }{15}-\tan \frac{\pi }{15}=\sqrt{3}+\sqrt{3},\tan \frac{6\pi }{15}\tan \frac{\pi }{15} $
$ \Rightarrow \tan \frac{6\pi }{15}-\tan \frac{\pi }{15}-\sqrt{3},\tan \frac{6\pi }{15}\tan \frac{\pi }{15}=\sqrt{3} $ .
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