A value of theta lying between 0 and frac{pi} | vedclass

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Question

(B) Let the given determinant be $\Delta$. Applying the column operation $C_1 	o C_1 + C_2$,we get: $\Delta = \left|\begin{array}{ccc} 1+\sin^2 	het

Vedclass · Accepted Answer

$\frac{7\pi}{24}$

Vedclass · Answer

(B) Let the given determinant be $\Delta$. Applying the column operation $C_1 	o C_1 + C_2$,we get: $\Delta = \left|\begin{array}{ccc} 1+\sin^2 	heta + \cos^2 	heta & \cos^2 	heta & 4\sin 4	heta \ \sin^2 	heta + 1 + \cos^2 	heta & 1+\cos^2 	heta & 4\sin 4	heta \ \sin^2 	heta + \cos^2 	heta & \cos^2 	heta & 1+4\sin 4	heta \end{array}ight| = 0$ Since $\sin^2 	heta + \cos^2 	heta = 1$,this simplifies to: $\Delta = \left|\begin{array}{ccc} 2 & \cos^2 	heta & 4\sin 4	heta \ 2 & 1+\cos^2 	heta & 4\sin 4	heta \ 1 & \cos^2 	heta & 1+4\sin 4	heta \end{array}ight| = 0$ Applying $R_1 	o R_1 - R_2$: $\Delta = \left|\begin{array}{ccc} 0 & -1 & 0 \ 2 & 1+\cos^2 	heta & 4\sin 4	heta \ 1 & \cos^2 	heta & 1+4\sin 4	heta \end{array}ight| = 0$ Expanding along the first row: $-(-1) 	imes [2(1+4\sin 4	heta) - 4\sin 4	heta] = 0$ $1 	imes [2 + 8\sin 4	heta - 4\sin 4	heta] = 0$ $2 + 4\sin 4	heta = 0$ $\sin 4	heta = -\frac{1}{2}$ Since $0 The values of $4	heta$ for which $\sin 4	heta = -\frac{1}{2}$ are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$. If $4	heta = \frac{7\pi}{6}$,then $	heta = \frac{7\pi}{24}$. If $4	heta = \frac{11\pi}{6}$,then $	heta = \frac{11\pi}{24}$. Comparing with the options,$\frac{7\pi}{24}$ is the correct value.

$A$ value of $\theta$ lying between $0$ and $\frac{\pi}{2}$ and satisfying $\left|\begin{array}{ccc} 1+\sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1+\cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1+4\sin 4\theta \end{array}\right| = 0$ is:

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