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
$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}}), x<\pi$

$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}})$

$=\tan ^{-1}(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}})$

$=\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$

$=\tan ^{-1}\left(\tan \frac{x}{2}\right)$

$=\frac{x}{2}$

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