CBSE Class 12 Mathematics 2025: question 22

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Step-by-step solution

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1. Since $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ and $\frac{\pi }{6}\in [0,\pi ]$, we have ${\cos }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}$. So the expression becomes ${\tan }^{-1}\left[2\sin \left(2\times \frac{\pi }{6}\right)\right]={\tan }^{-1}\left[2\sin \frac{\pi }{3}\right]$.

   1 mark

2. Now $2\sin \frac{\pi }{3}=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$. Since $\frac{\pi }{3}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, ${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$.

   1 mark

Answer: $\frac{\pi }{3}$

Revise the chapter

• Chapter: Inverse Trigonometric Functions (notes, formulas, where marks are lost and every past question)