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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

Considering the principal values of the inverse trigonometric functions, \(\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2} \lt x \lt \frac{1}{\sqrt{2}}\), is equal to

  1. A \(\frac{\pi}{4}+\sin ^{-1} x\)
  2. B \(\frac{\pi}{6}+\sin ^{-1} x\)
  3. C \(\frac{-5 \pi}{6}-\sin ^{-1} x\)
  4. D \(\frac{5 \pi}{6}-\sin ^{-1} x\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\pi}{6}+\sin ^{-1} x\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right), \frac{-1}{2} \lt x \lt \frac{1}{\sqrt{2}} \\ & \Rightarrow \text { Let } \sin ^{-1}(x)=\theta \quad \frac{-\pi}{6} \lt \theta \lt \frac{\pi}{4} \\ & \Rightarrow x=\sin \theta, \text { then }…