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)

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