begin{vmatrix} cos^2theta & -sin^2theta sin^ | vedclass

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(B) The determinant of the matrix is calculated as: $\begin{vmatrix} \cos^2	heta & -\sin^2	heta \ \sin^2	heta & \cos^2	heta \end{vmatrix} = (\cos

Vedclass · Accepted Answer

$\frac{1}{4}(3 + \cos 4	heta)$

Vedclass · Answer

(B) The determinant of the matrix is calculated as: $\begin{vmatrix} \cos^2	heta & -\sin^2	heta \ \sin^2	heta & \cos^2	heta \end{vmatrix} = (\cos^2	heta)(\cos^2	heta) - (-\sin^2	heta)(\sin^2	heta) = \cos^4	heta + \sin^4	heta$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we get: $\cos^4	heta + \sin^4	heta = (\cos^2	heta + \sin^2	heta)^2 - 2\sin^2	heta\cos^2	heta = 1 - 2\sin^2	heta\cos^2	heta$. Since $\sin 2	heta = 2\sin	heta\cos	heta$, we have $\sin^2 2	heta = 4\sin^2	heta\cos^2	heta$, so $2\sin^2	heta\cos^2	heta = \frac{1}{2}\sin^2 2	heta$. Thus, the expression becomes $1 - \frac{1}{2}\sin^2 2	heta$. Using the identity $\sin^2 2	heta = \frac{1 - \cos 4	heta}{2}$, we substitute: $1 - \frac{1}{2}(\frac{1 - \cos 4	heta}{2}) = 1 - \frac{1}{4} + \frac{1}{4}\cos 4	heta = \frac{3}{4} + \frac{1}{4}\cos 4	heta = \frac{1}{4}(3 + \cos 4	heta)$. Therefore, option $(B)$ is correct.

$\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} = \dots$

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