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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
$\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right]$ is equal to
  • A
    1
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{4}$
✓

Answer

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

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