MCQ
$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$ बराबर है
  • A
    $\frac{1}{3}$
  • B
    1
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{2}$

Answer

$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$= $\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\sin \frac{\pi}{6}\right)\right]$ $\left(\because \sin \frac{\pi}{6}=\frac{1}{2}\right)$
= $\sin \left[\frac{\pi}{3}-\sin ^{-1}\left\{\sin \left(-\frac{\pi}{6}\right)\right\}\right]$ $[\because \sin (-x)=-\sin x]$
= $\sin \left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]$ = $\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)$ = $\sin \frac{\pi}{2}$ = 1
 

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