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CBSE Board Examination 2025 · Section B · 2 marks

Q22: Evaluate tan⁡−1[2sin⁡(2cos⁡−132)]\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt3}{2}\right)\right]

Board 2025Ch 2 · Inverse Trigonometric Functions
The question

Evaluate : tan⁡−1[2sin⁡(2cos⁡−132)]\tan^{-1}\left[2\sin\left(2\cos^{-1}\dfrac{\sqrt3}{2}\right)\right]

Exam answer

cos⁡−132=π6\cos^{-1}\frac{\sqrt3}{2} = \frac{\pi}{6}⁠, as π6∈[0,π]\frac{\pi}{6} \in [0, \pi] and cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt3}{2}⁠.

Why?

Work from the inside out. The innermost part is cos⁡−132\cos^{-1}\frac{\sqrt3}{2}⁠.

cos⁡−1\cos^{-1} always gives its principal value, the one angle in [0,π][0, \pi] with the given cosine.

Watch out: −π6-\frac{\pi}{6} also has cosine 32\frac{\sqrt3}{2}⁠, but it is not in [0,π][0, \pi]⁠.

tan⁡−1[2sin⁡(2cos⁡−132)]=tan⁡−1[2sin⁡(2×π6)]=tan⁡−1[2sin⁡π3]=tan⁡−1[2×32]=tan⁡−13\begin{aligned} &\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt3}{2}\right)\right] \\ &= \tan^{-1}\left[2\sin\left(2\times\frac{\pi}{6}\right)\right] \\ &= \tan^{-1}\left[2\sin\frac{\pi}{3}\right] \\ &= \tan^{-1}\left[2\times\frac{\sqrt3}{2}\right] \\ &= \tan^{-1}\sqrt3 \end{aligned}
Why?

Put π6\frac{\pi}{6} in place of cos⁡−132\cos^{-1}\frac{\sqrt3}{2}⁠. Then simplify one layer at a time:

2×π6=π3sin⁡π3=322×32=3\begin{aligned} 2\times\frac{\pi}{6} &= \frac{\pi}{3} \\ \sin\frac{\pi}{3} &= \frac{\sqrt3}{2} \\ 2\times\frac{\sqrt3}{2} &= \sqrt3 \end{aligned}

tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠, as π3∈(−π2,π2)\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) and tan⁡π3=3\tan\frac{\pi}{3} = \sqrt3⁠.

Why?

The principal value branch of tan⁡−1\tan^{-1} is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠. The only angle there whose tangent is 3\sqrt3 is π3\frac{\pi}{3}⁠.

Watch out: 4π3\frac{4\pi}{3} also has tangent 3\sqrt3⁠, but it is outside the branch.

∴\therefore The value is π3\dfrac{\pi}{3}⁠.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Inverse trigonometric functions: principal values.

Question type: A trig function of an inverse

How to spot it: click to see

Type 6. A trig function of an inverse: work from the inside out

Spot it: an inverse sitting inside sin, cos or tan, like cos⁡(2sin⁡−112)\cos\left(2\sin^{-1}\frac12\right)⁠.

Here: The innermost inverse is cos⁡−132=π6\cos^{-1}\frac{\sqrt3}{2} = \frac{\pi}{6}⁠. Then 2sin⁡π3=32\sin\frac{\pi}{3} = \sqrt3⁠, and tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠.

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