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STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 2. inverse trigonometric function1 Mark
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
(a) ${\tan ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right] = {\tan ^{ - 1}}\left[ {\frac{{\sin \,(\pi /2 - x)}}{{1 + \cos \,(\pi /2 - x)}}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{2\,\sin \,(\pi /4 - x/2)\,\cos \,(\pi /4 - x/2)}}{{2\,{{\cos }^2}\,(\pi /4 - x/2)}}} \right]$

$ = {\tan ^{ - 1}}\tan \,\left( {\frac{\pi }{4} - \frac{x}{2}} \right) = \frac{\pi }{4} - \frac{x}{2}$.

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