Considering the principal values of the inverse trigonometric functions, $\sin ^{-1}\left(\frac{\sqrt{3}}{2}…

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. $\frac{\pi}{4}+\sin ^{-1} x$
  2. $\frac{\pi}{6}+\sin ^{-1} x$
  3. $\frac{-5 \pi}{6}-\sin ^{-1} x$
  4. $\frac{5 \pi}{6}-\sin ^{-1} x$

Solution

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

Asked in: JEE Main 2025 (04 Apr Shift 1)

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