Question

If $x \in (7\pi , 8 \pi),$ then $\tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} = $

• A. $ - \frac{x}{2}$
• B. $ \frac{x}{2}$
• C. $4 \pi - \frac{x}{2}$
• D. None of these

Answer 1

Answer: (C)

$\tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} =\tan^{-1}\left(-\tan \frac{x}{2}\right) $
$= \tan^{-1} \tan\left(4 \pi- \frac{\pi}{2}\right)= 4\pi - \frac{x}{2}$
$ \left[\because 7 \pi< x < 8\pi \Rightarrow \frac{7\pi}{2} < \frac{x}{2} < 4 \pi, \text{so} \tan \frac{x}{2} < 0\right] $