The value of theta lying between theta = 0 an | vedclass

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(D) Let the determinant be $\Delta$. We apply the column operation $C_1 	o C_1 + C_2$: $\Delta = \left| \begin{array}{ccc} 1 + \sin^2 	heta + \cos^2

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both $(A)$ and $(C)$

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(D) Let the determinant be $\Delta$. We apply the column operation $C_1 	o C_1 + C_2$: $\Delta = \left| \begin{array}{ccc} 1 + \sin^2 	heta + \cos^2 	heta & \cos^2 	heta & 4 \sin 4 	heta \ \sin^2 	heta + 1 + \cos^2 	heta & 1 + \cos^2 	heta & 4 \sin 4 	heta \ \sin^2 	heta + \cos^2 	heta & \cos^2 	heta & 1 + 4 \sin 4 	heta \end{array} ight| = 0$ Since $\sin^2 	heta + \cos^2 	heta = 1$,this simplifies to: $\Delta = \left| \begin{array}{ccc} 2 & \cos^2 	heta & 4 \sin 4 	heta \ 2 & 1 + \cos^2 	heta & 4 \sin 4 	heta \ 1 & \cos^2 	heta & 1 + 4 \sin 4 	heta \end{array} ight| = 0$ Apply $R_2 	o R_2 - R_1$: $\Delta = \left| \begin{array}{ccc} 2 & \cos^2 	heta & 4 \sin 4 	heta \ 0 & 1 & 0 \ 1 & \cos^2 	heta & 1 + 4 \sin 4 	heta \end{array} ight| = 0$ Expanding along $R_2$: $1 	imes [2(1 + 4 \sin 4 	heta) - 4 \sin 4 	heta] = 0$ $2 + 8 \sin 4 	heta - 4 \sin 4 	heta = 0$ $2 + 4 \sin 4 	heta = 0 \implies \sin 4 	heta = -1/2$ Since $0 The values of $4 	heta$ for which $\sin 4 	heta = -1/2$ are $7\pi/6$ and $11\pi/6$. Thus,$	heta = 7\pi/24$ and $	heta = 11\pi/24$. Both values lie in the range $(0, \pi/2)$.

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