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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
$\sin \left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right] = $
  • A
    $\frac{{\sqrt 3 }}{2}$
  • B
    $ - \frac{{\sqrt 3 }}{2}$
  • ✓
    $\frac{1}{2}$
  • D
    $ - \frac{1}{2}$
✓

Answer

Correct option: C.
$\frac{1}{2}$
c
(c) $\sin \,\left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right] = \cos \,{\sin ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right)$

$ = \cos \,{\cos ^{ - 1}}\sqrt {1 - \frac{3}{4}} = \frac{1}{2}$.

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