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Matrices & Determinants: Vmatrix Vmatrix Equals

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

Question
If 0<θ<π20 < \theta < \dfrac{\pi}{2} and ∣1+sin⁡2θcos⁡2θ4sin⁡4θsin⁡2θ1+cos⁡2θ4sin⁡4θsin⁡2θcos⁡2θ1+4sin⁡4θ∣=0\begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta \\ \sin^2\theta & 1+\cos^2\theta & 4\sin4\theta \\ \sin^2\theta & \cos^2\theta & 1+4\sin4\theta \end{vmatrix} = 0, then θ\theta equals:
Aπ24, 5π24\dfrac{\pi}{24},\, \dfrac{5\pi}{24}
B5π24, 7π24\dfrac{5\pi}{24},\, \dfrac{7\pi}{24}
C7π24, 11π24\dfrac{7\pi}{24},\, \dfrac{11\pi}{24}correct
DNone of these
Solution
Step 1: Apply R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1
∣1+sin⁡2θcos⁡2θ4sin⁡4θ−110−101∣=0\begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = 0
 Step 2: Expand along row 1 using cofactors
A11=(+1)∣1001∣=+1,A12=(−1)1+2∣−10−11∣=(−1)(−1)=+1A_{11} = (+1)\begin{vmatrix}1&0\\0&1\end{vmatrix} = +1, \quad A_{12} = (-1)^{1+2}\begin{vmatrix}-1&0\\-1&1\end{vmatrix} = (-1)(-1) = +1
A13=(+1)∣−11−10∣=+1\quad A_{13} = (+1)\begin{vmatrix}-1&1\\-1&0\end{vmatrix} = +1
 Therefore:
(1+sin⁡2θ)(+1)+cos⁡2θ(+1)+4sin⁡4θ(+1)=0(1+\sin^2\theta)(+1) + \cos^2\theta(+1) + 4\sin4\theta(+1) = 0
1+sin⁡2θ+cos⁡2θ+4sin⁡4θ=0  ⟹  2+4sin⁡4θ=01 + \sin^2\theta + \cos^2\theta + 4\sin4\theta = 0 \implies 2 + 4\sin4\theta = 0
 Step 3: Solve for θ\theta
sin⁡4θ=−12  ⟹  4θ=nπ+(−1)n ⁣(−π6)\sin4\theta = -\frac{1}{2} \implies 4\theta = n\pi+(-1)^n\!\left(-\frac{\pi}{6}\right)
 For θ∈(0,π2)\theta \in \left(0,\dfrac{\pi}{2}\right): n=1n=1: 4θ=7π6  ⟹  θ=7π244\theta = \dfrac{7\pi}{6} \implies \theta = \dfrac{7\pi}{24}. n=2\quad n=2: 4θ=11π6  ⟹  θ=11π244\theta = \dfrac{11\pi}{6} \implies \theta = \dfrac{11\pi}{24}. Answer: (3)
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