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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
$\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=$
  • ✓
    $\frac{\pi}{4}-\frac{x}{2}$
  • B
    $\frac{\pi}{4}+\frac{x}{2}$
  • C
    $\frac{x}{2}$
  • D
    $\frac{\pi}{4}-x$
✓

Answer

Correct option: A.
$\frac{\pi}{4}-\frac{x}{2}$
(A) $\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan ^{-1}\left[\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right]$
$=\tan ^{-1}\left[\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]=\frac{\pi}{4}-\frac{x}{2}$

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