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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
Find the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.
  • ✓
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{6}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{2}$
✓

Answer

Correct option: A.
$\frac{\pi}{4}$
(a) : Let $x=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$, then $\sin x=\frac{1}{\sqrt{2}}$
We know that the range of principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and $\sin x=\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \Rightarrow x=\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \quad$ Principal value is $\frac{\pi}{4}$.

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