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Matrices & Determinants: Vmatrix Vmatrix

JEE Maths question with a full step-by-step solution.

Question
If f(θ)=∣cos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θsin⁡θsin⁡2θcos⁡θsin⁡θ−cos⁡θ0∣f(\theta) = \begin{vmatrix} \cos^2\theta & \cos\theta\sin\theta & -\sin\theta \\ \cos\theta\sin\theta & \sin^2\theta & \cos\theta \\ \sin\theta & -\cos\theta & 0 \end{vmatrix}, then:
Af ⁣(π4)=1f\!\left(\dfrac{\pi}{4}\right) = 1correct
Bf ⁣(π4)=2f\!\left(\dfrac{\pi}{4}\right) = 2
Cf ⁣(π8)=3f\!\left(\dfrac{\pi}{8}\right) = 3
DNone of these
Solution
Step 1: Apply column operations C1→C1−sin⁡θ C3C_1 \to C_1 - \sin\theta\, C_3 and C2→C2+cos⁡θ C3C_2 \to C_2 + \cos\theta\, C_3
f(θ)=∣10−sin⁡θ01cos⁡θsin⁡θ−cos⁡θ0∣f(\theta) = \begin{vmatrix} 1 & 0 & -\sin\theta \\ 0 & 1 & \cos\theta \\ \sin\theta & -\cos\theta & 0 \end{vmatrix}
 Step 2: Apply R3→R3−sin⁡θ R1+cos⁡θ R2R_3 \to R_3 - \sin\theta\, R_1 + \cos\theta\, R_2
=∣10−sin⁡θ01cos⁡θ001∣=1= \begin{vmatrix} 1 & 0 & -\sin\theta \\ 0 & 1 & \cos\theta \\ 0 & 0 & 1 \end{vmatrix} = 1
 The determinant equals 11 for all θ\theta, so f ⁣(π4)=1f\!\left(\dfrac{\pi}{4}\right) = 1. Answer: (1)
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