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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 $\tan ^{-1}(-\sqrt{3})$
  • A
    $\frac{\pi}{3}$
  • ✓
    $-\frac{\pi}{3}$
  • C
    $-\frac{\pi}{6}$
  • D
    $\frac{5\pi}{6}$
✓

Answer

Correct option: B.
$-\frac{\pi}{3}$
b
Let $\tan ^{-1}(-\sqrt{3})=y$

Then, $\tan y=-\sqrt{3}=-\tan \frac{\pi}{3}=\tan \left(-\frac{\pi}{3}\right)$

We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2} \frac{\pi}{2}\right)$ and $\tan \left(-\frac{\pi}{3}\right)$ is $-\sqrt{3}$

Therefore, known that the principal value of $\tan ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$

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