Question

If $\theta =\frac{17\pi }{3}$ then, $tan\theta -cot\theta =\dots$

• A. $\frac{1}{2\sqrt{3}}$
• B. $\frac{-1}{2\sqrt{3}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $-\frac{2}{\sqrt{3}}$

Answer 1

Answer: (D)

We have, $\theta =\frac{17\pi }{3}$
$\therefore \tan \theta -\cot \theta =\tan \left(\frac{17\pi }{3}\right)-\cot \left(\frac{17\pi }{3}\right)$
$=\tan \left(5\pi +\frac{2\pi }{3}\right)-\cot \left(5\pi +\frac{2\pi }{3}\right)$
$=\tan \frac{2\pi }{3}-\cot \frac{2\pi }{3}$
$=\tan \left(\pi -\frac{\pi }{3}\right)-\cot \left(\pi -\frac{\pi }{3}\right)$
$=-\tan \pi /3+\cot \pi /3$
$=-\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{-3+1}{\sqrt{3}}=\frac{-2}{\sqrt{3}}$