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\( \Large \tan \frac{2 \pi }{5} - \tan \frac{ \pi }{15} - \sqrt{3} \tan \frac{2 \pi }{5} \tan \frac{ \pi }{15} \) is equal to:

A) \( \Large -\sqrt{3} \)
B) \( \Large \frac{1}{\sqrt{3}} \)
C) \( \Large 1 \)
D) \( \Large \sqrt{3} \)

Correct Answer:




D) \( \Large \sqrt{3} \)

Description for Correct answer:
Now, \( \Large \tan \frac{ \pi }{3}=\tan \left(\frac{6 \pi }{15}-\frac{ \pi }{15}\right)= \frac{\tan\frac{6 \pi }{15}-\tan\frac{ \pi }{15}}{1+\tan\frac{6 \pi }{15}\tan\frac{ \pi }{15}} \)

=> \( \Large \tan\frac{6 \pi }{15}-\tan\frac{ \pi }{15}=\sqrt{3}+\sqrt{3}\tan\frac{6 \pi }{15}\tan\frac{ \pi }{15} \)

=> \( \Large \tan\frac{6 \pi }{15}-\tan\frac{ \pi }{15}-\sqrt{3}\tan\frac{6 \pi }{15}\tan\frac{ \pi }{15}=\sqrt{3} \)

=> \( \Large \tan\frac{2 \pi }{5}-\tan\frac{ \pi }{15}-\sqrt{3}\tan\frac{2 \pi }{5}\tan\frac{ \pi }{15}=\sqrt{3} \)

Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio

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