If frac{2+3i sin theta}{1-2i sin theta} is pu | vedclass

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(D) Let $z = \frac{2+3i \sin 	heta}{1-2i \sin 	heta}$.To make $z$ purely imaginary,the real part of $z$ must be zero.Multiply the numerator and deno

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$\frac{2}{3}$

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(D) Let $z = \frac{2+3i \sin 	heta}{1-2i \sin 	heta}$.To make $z$ purely imaginary,the real part of $z$ must be zero.Multiply the numerator and denominator by the conjugate of the denominator: $1+2i \sin 	heta$.$z = \frac{(2+3i \sin 	heta)(1+2i \sin 	heta)}{(1-2i \sin 	heta)(1+2i \sin 	heta)} = \frac{2 + 4i \sin 	heta + 3i \sin 	heta + 6i^2 \sin^2 	heta}{1 + 4 \sin^2 	heta}$.Since $i^2 = -1$,$z = \frac{2 - 6 \sin^2 	heta + 7i \sin 	heta}{1 + 4 \sin^2 	heta}$.The real part is $\frac{2 - 6 \sin^2 	heta}{1 + 4 \sin^2 	heta}$.Setting the real part to $0$: $2 - 6 \sin^2 	heta = 0 \implies 6 \sin^2 	heta = 2 \implies \sin^2 	heta = \frac{1}{3}$.Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get $\cos^2 	heta = 1 - \frac{1}{3} = \frac{2}{3}$.

If $\frac{2+3i \sin \theta}{1-2i \sin \theta}$ is purely imaginary,then $\cos^2 \theta=$

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