MCQ
Write the function in the simplest form: $\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}}), x<\pi$
- A$2x$
- ✓$\frac{x}{2}$
- C$\frac{\pi}{2}$
- D$\pi$
✓
Answer
Correct option: B.
$\frac{x}{2}$
b
$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}}), x<\pi$
$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}}), x<\pi$
$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}})$
$=\tan ^{-1}(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}})$
$=\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$
$=\tan ^{-1}\left(\tan \frac{x}{2}\right)$
$=\frac{x}{2}$
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