If left|begin{array}{cc}sin ^2 theta & cos ^2 | vedclass

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(B) To evaluate the determinant $\left|\begin{array}{cc}\sin ^2 	heta & \cos ^2 	heta \ -\cos ^2 	heta & \sin ^2 	heta\end{array}ight|$,we use

Vedclass · Accepted Answer

$\sin^4 	heta + \cos^4 	heta$

Vedclass · Answer

(B) To evaluate the determinant $\left|\begin{array}{cc}\sin ^2 	heta & \cos ^2 	heta \ -\cos ^2 	heta & \sin ^2 	heta\end{array}ight|$,we use the formula for a $2 	imes 2$ determinant: $\left|\begin{array}{cc}a & b \ c & d\end{array}ight| = ad - bc$. Applying this,we get: $(\sin^2 	heta)(\sin^2 	heta) - (\cos^2 	heta)(-\cos^2 	heta)$ $= \sin^4 	heta + \cos^4 	heta$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,we can write: $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta$. Since $\sin^2 	heta + \cos^2 	heta = 1$,this becomes: $1^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2}(4 \sin^2 	heta \cos^2 	heta) = 1 - \frac{1}{2}(2 \sin 	heta \cos 	heta)^2$. Using the double angle identity $\sin 2 	heta = 2 \sin 	heta \cos 	heta$,we get: $1 - \frac{1}{2} \sin^2 2 	heta$.

If $\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right| = $ . . . . . . .

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