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CBSE Sample Paper 2025-26 · Section A · 1 mark

Q6: Determinant of cos⁡67∘\cos 67^\circ⁠, sin⁡67∘\sin 67^\circ⁠, sin⁡23∘\sin 23^\circ⁠, cos⁡23∘\cos 23^\circ

Sample 2025-26Ch 4 · Determinants
The question

Value of the determinant ∣cos⁡67∘sin⁡67∘sin⁡23∘cos⁡23∘∣\begin{vmatrix} \cos 67^\circ & \sin 67^\circ \\ \sin 23^\circ & \cos 23^\circ \end{vmatrix} is

(A) 00

(B) 12\frac{1}{2}

(C) 32\frac{\sqrt{3}}{2}

(D) 11

Exam answer

Expanding the determinant,

Δ=cos⁡67∘cos⁡23∘−sin⁡67∘sin⁡23∘\begin{aligned} \Delta &= \cos 67^\circ\cos 23^\circ \\ &\quad - \sin 67^\circ\sin 23^\circ \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⁡67∘a = \cos 67^\circ⁠, b=sin⁡67∘b = \sin 67^\circ⁠, c=sin⁡23∘c = \sin 23^\circ and d=cos⁡23∘d = \cos 23^\circ⁠.

Δ=cos⁡(67∘+23∘)=cos⁡90∘=0\begin{aligned} \Delta &= \cos(67^\circ + 23^\circ) \\ &= \cos 90^\circ = 0 \end{aligned}
Why?

The compound angle formula:

cos⁡Acos⁡B−sin⁡Asin⁡B=cos⁡(A+B)\begin{aligned} &\cos A\cos B - \sin A\sin B \\ &= \cos(A + B) \end{aligned}

With A=67∘A = 67^\circ and B=23∘B = 23^\circ⁠, A+B=90∘A + B = 90^\circ⁠, and cos⁡90∘=0\cos 90^\circ = 0⁠.

Another way: 67° and 23° are complementary, so cos⁡23∘=sin⁡67∘\cos 23^\circ = \sin 67^\circ and sin⁡23∘=cos⁡67∘\sin 23^\circ = \cos 67^\circ⁠. The two products are then equal, and their difference is 0.

Answer: (A) 00

Where this comes from

Chapter 4, Determinants: Value of a 2 × 2 determinant.

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: cross-multiply to get cos⁡67∘cos⁡23∘−sin⁡67∘sin⁡23∘\cos 67^\circ\cos 23^\circ - \sin 67^\circ\sin 23^\circ⁠, which is cos⁡90∘=0\cos 90^\circ = 0⁠.

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