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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 $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
  • A
    $\frac{2 \pi}{3}$
  • B
    $\frac{\pi}{3}$
  • ✓
    $\frac{5 \pi}{6}$
  • D
    $\frac{\pi}{4}$
✓

Answer

Correct option: C.
$\frac{5 \pi}{6}$
(c) : Let $x=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$, then $\cos x=\frac{-\sqrt{3}}{2}$
We know that the range of principal value branch of $\cos ^{-1}$ is $[0, \pi]$
$\Rightarrow \cos x=\cos \left(\pi-\frac{\pi}{6}\right)=\frac{-\sqrt{3}}{2} \Rightarrow x=\frac{5 \pi}{6} \in[0, \pi]$
$\therefore \quad$ Principal value is $\frac{5 \pi}{6}$.

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