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CBSE Board Examination 2026 · Section A · 1 mark

Q5: A value of xx making a trigonometric determinant equal 1

Board 2026Ch 4 · Determinants
The question

One of the values of xx for which ∣cos⁡xsin⁡x−cos⁡xsin⁡x∣=1\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1 is

(A) 00

(B) π4\frac{\pi}{4}

(C) π3\frac{\pi}{3}

(D) π2\frac{\pi}{2}

Exam answer

Expanding the determinant,

cos⁡xsin⁡x−sin⁡x(−cos⁡x)=1⇒ cos⁡xsin⁡x+sin⁡xcos⁡x=1\begin{aligned} &\cos x\sin x - \sin x(-\cos x) = 1 \\ \Rightarrow\ &\cos x\sin x + \sin x\cos x = 1 \end{aligned}
Why?

A 2×22 \times 2 determinant is

∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Here a=cos⁡xa = \cos x⁠, b=sin⁡xb = \sin x⁠, c=−cos⁡xc = -\cos x and d=sin⁡xd = \sin x⁠.

Watch out: bc=sin⁡x(−cos⁡x)bc = \sin x(-\cos x) is negative, and subtracting it adds.

⇒ 2sin⁡xcos⁡x=1⇒ sin⁡2x=1⇒ 2x=π2, so x=π4\begin{aligned} \Rightarrow\ &2\sin x\cos x = 1 \\ \Rightarrow\ &\sin 2x = 1 \\ \Rightarrow\ &2x = \frac{\pi}{2},\ \text{so } x = \frac{\pi}{4} \end{aligned}
Why?

The double angle formula:

sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x

First, cos⁡xsin⁡x\cos x\sin x and sin⁡xcos⁡x\sin x\cos x are the same product, so

cos⁡xsin⁡x+sin⁡xcos⁡x=2sin⁡xcos⁡x=sin⁡2x\begin{aligned} &\cos x\sin x + \sin x\cos x \\ &= 2\sin x\cos x = \sin 2x \end{aligned}

Among positive angles, the sine first equals 1 at π2\frac{\pi}{2}⁠, so take 2x=π22x = \frac{\pi}{2}⁠. The question asks for one of the values: 2x=5π22x = \frac{5\pi}{2} also works, giving x=5π4x = \frac{5\pi}{4}⁠, but that is not an option.

2x=π2x=π2÷2=π4\begin{aligned} 2x &= \frac{\pi}{2} \\ x &= \frac{\pi}{2} \div 2 = \frac{\pi}{4} \end{aligned}

Check: at x=0,π3,π2x = 0, \frac{\pi}{3}, \frac{\pi}{2}⁠, sin⁡2x=0,32,0\sin 2x = 0, \frac{\sqrt{3}}{2}, 0⁠, not 1.

Answer: (B) π4\dfrac{\pi}{4}

Where this comes from

Chapter 4, Determinants: Evaluating 2 × 2 determinants.

Question type: Evaluate a determinant

How to spot it: click to see

Type 1. Evaluate a determinant: cross for 2×22\times2⁠, expand for 3×33\times3

Spot it: "evaluate" and a square block of numbers between straight bars.

Here: The bars hold a 2×22 \times 2 in cos⁡x\cos x and sin⁡x\sin x⁠: cross-multiply to get 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x⁠, set it equal to 1, and x=π4x = \frac{\pi}{4}⁠.

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