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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\tan \frac{2 \pi}{5}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{2 \pi}{5} \tan \frac{\pi}{15}$ is equal to
  • A
    $-\sqrt{3}$
  • B
    $\frac{1}{\sqrt{3}}$
  • C
    1
  • ✓
    $\sqrt{3}$
✓

Answer

Correct option: D.
$\sqrt{3}$
(D)
$\tan \left(\frac{6 \pi}{15}-\frac{\pi}{15}\right)-\tan \frac{\pi}{3}$
$\Rightarrow \frac{\tan \frac{6 \pi}{15}-\tan \frac{\pi}{15}}{1+\tan \frac{6 \pi}{15} \tan \frac{\pi}{15}}=\tan \frac{\pi}{3}$
$\Rightarrow \tan \frac{6 \pi}{15}-\tan \frac{\pi}{15}=\sqrt{3}+\sqrt{3} \tan \frac{6 \pi}{15} \tan \frac{\pi}{15}$
$\Rightarrow \tan \frac{6 \pi}{15}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{6 \pi}{15} \tan \frac{\pi}{15}=\sqrt{3}$
$\Rightarrow \tan \frac{2 \pi}{5}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{2 \pi}{5} \tan \frac{\pi}{15}=\sqrt{3}$

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