In any triangle ABC,the expression {sin ^2}fr | vedclass

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(C) We know that in a triangle $ABC$,$A+B+C = 180^\circ$,so $\frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2}$.Using the identity $\sin^2 	heta = \

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$1 - 2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

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(C) We know that in a triangle $ABC$,$A+B+C = 180^\circ$,so $\frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2}$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have:${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2} = \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \sin^2 \frac{C}{2}$$= 1 - \frac{1}{2}(\cos A + \cos B) + \sin^2 \frac{C}{2}$$= 1 - \cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) + \sin^2 \frac{C}{2}$Since $\cos(\frac{A+B}{2}) = \sin \frac{C}{2}$,we get:$= 1 - \sin \frac{C}{2} \cos(\frac{A-B}{2}) + \sin^2 \frac{C}{2}$$= 1 - \sin \frac{C}{2} [\cos(\frac{A-B}{2}) - \sin \frac{C}{2}]$$= 1 - \sin \frac{C}{2} [\cos(\frac{A-B}{2}) - \cos(\frac{A+B}{2})]$Using $\cos X - \cos Y = 2\sin(\frac{X+Y}{2})\sin(\frac{Y-X}{2})$,we get:$= 1 - \sin \frac{C}{2} [2\sin \frac{A}{2} \sin \frac{B}{2}]$$= 1 - 2\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

In any triangle $ABC$,the expression ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

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