The value of theta lying between -frac{pi}{4} | vedclass

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(D) Given the determinant $D = \begin{vmatrix} 1 + \sin^2 A & \cos^2 A & 2 \sin 4	heta \ \sin^2 A & 1 + \cos^2 A & 2 \sin 4	heta \ \sin^2 A & \cos

Vedclass · Accepted Answer

All of the above

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(D) Given the determinant $D = \begin{vmatrix} 1 + \sin^2 A & \cos^2 A & 2 \sin 4	heta \ \sin^2 A & 1 + \cos^2 A & 2 \sin 4	heta \ \sin^2 A & \cos^2 A & 1 + 2 \sin 4	heta \end{vmatrix} = 0$.Applying $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$,we get:$D = \begin{vmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ \sin^2 A & \cos^2 A & 1 + 2 \sin 4	heta \end{vmatrix} = 0$.Expanding along the first row:$1(1 + 2 \sin 4	heta + \cos^2 A) + 1(0 + \sin^2 A) = 0$.$1 + 2 \sin 4	heta + \cos^2 A + \sin^2 A = 0$.Since $\sin^2 A + \cos^2 A = 1$,we have $1 + 2 \sin 4	heta + 1 = 0$,which simplifies to $2 + 2 \sin 4	heta = 0$.Thus,$\sin 4	heta = -1$.The general solution for $4	heta = 2n\pi - \frac{\pi}{2}$ is $	heta = \frac{n\pi}{2} - \frac{\pi}{8}$.For the range $-\frac{\pi}{4} Since the equation is independent of $A$,any $A \in [0, \frac{\pi}{2}]$ satisfies the equation. Thus,all given options are correct.

The value of $\theta$ lying between $-\frac{\pi}{4}$ and $\frac{\pi}{2}$ and $0 \le A \le \frac{\pi}{2}$ satisfying the equation $\begin{vmatrix} 1 + \sin^2 A & \cos^2 A & 2 \sin 4\theta \\ \sin^2 A & 1 + \cos^2 A & 2 \sin 4\theta \\ \sin^2 A & \cos^2 A & 1 + 2 \sin 4\theta \end{vmatrix} = 0$ are:

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