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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
Find the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
  • ✓
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{2}$
✓

Answer

Correct option: A.
$\frac{\pi}{6}$
a
Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=y \cdot$ Then, $\sec y=\frac{2}{\sqrt{3}}=\sec \left(\frac{\pi}{6}\right)$

We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$ and $\sec \left(\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}}$

Therefore, the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$

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