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CBSE Sample Paper 2026-27 · Section A · 1 mark

Q3: Value of cot⁡−1(2cos⁡(12sin⁡−132))\cot^{-1}\left(2\cos\left(\frac12\sin^{-1}\frac{\sqrt3}{2}\right)\right)

Sample 2026-27Ch 2 · Inverse Trigonometric Functions
The question

Value of cot⁡−1(2cos⁡(12sin⁡−1(32)))\cot^{-1}\left(2\cos\left(\frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right)

(A) π6\frac{\pi}{6}

(B) π4\frac{\pi}{4}

(C) π3\frac{\pi}{3}

(D) π2\frac{\pi}{2}

Exam answer

sin⁡−1(32)=π3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}⁠, so the value is

Why?

Work from the inside out. The principal value of sin⁡−1\sin^{-1} lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, and

sin⁡π3=32\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}

with π3\frac{\pi}{3} inside that interval. Watch out: 2π3\frac{2\pi}{3} has the same sine, but it is not a principal value.

cot⁡−1(2cos⁡(12×π3))=cot⁡−1(2cos⁡π6)=cot⁡−1(2×32)=cot⁡−1(3)\begin{aligned} &\cot^{-1}\left(2\cos\left(\frac{1}{2} \times \frac{\pi}{3}\right)\right) \\ &= \cot^{-1}\left(2\cos\frac{\pi}{6}\right) \\ &= \cot^{-1}\left(2 \times \frac{\sqrt{3}}{2}\right) \\ &= \cot^{-1}(\sqrt{3}) \end{aligned}
Why?

Half of π3\frac{\pi}{3} is π6\frac{\pi}{6}⁠, and cos⁡π6=32\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}⁠. The 2 cancels the denominator 2, leaving 3\sqrt{3}⁠.

12×π3=1×π2×3=π62×32=232=3\begin{aligned} \frac{1}{2} \times \frac{\pi}{3} &= \frac{1 \times \pi}{2 \times 3} = \frac{\pi}{6} \\[4pt] 2 \times \frac{\sqrt{3}}{2} &= \frac{2\sqrt{3}}{2} = \sqrt{3} \end{aligned}
cot⁡−1(3)=π6\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}
Why?

The principal value of cot⁡−1\cot^{-1} lies in (0,π)(0, \pi)⁠. Using cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}⁠,

cot⁡π6=cos⁡(π/6)sin⁡(π/6)=3/21/2=32×21=3\begin{aligned} \cot\frac{\pi}{6} &= \frac{\cos(\pi/6)}{\sin(\pi/6)} \\ &= \frac{\sqrt{3}/2}{1/2} \\ &= \frac{\sqrt{3}}{2} \times \frac{2}{1} \\ &= \sqrt{3} \end{aligned}

Since π6\frac{\pi}{6} is in (0,π)(0, \pi)⁠, the value is π6\frac{\pi}{6}⁠. Watch out: cot⁡π3=13\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}⁠, not 3\sqrt{3}⁠, so (C) is wrong.

Answer: (A) π6\dfrac{\pi}{6}

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Principal values of inverse trig functions.

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 sin⁡−132=π3\sin^{-1}\frac{\sqrt3}{2} = \frac{\pi}{3}⁠. Then 2cos⁡π6=32\cos\frac{\pi}{6} = \sqrt3⁠, and cot⁡−13=π6\cot^{-1}\sqrt3 = \frac{\pi}{6}⁠.

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