If A and B are supplementary angles,then sin^ | vedclass

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Question

(A) Given that $A$ and $B$ are supplementary angles,we have $A + B = 180^{\circ}$.This implies $B = 180^{\circ} - A$,so $\frac{B}{2} = 90^{\circ} - \f

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given that $A$ and $B$ are supplementary angles,we have $A + B = 180^{\circ}$.This implies $B = 180^{\circ} - A$,so $\frac{B}{2} = 90^{\circ} - \frac{A}{2}$.Substituting this into the expression:$\sin^{2} \frac{A}{2} + \sin^{2} \frac{B}{2} = \sin^{2} \frac{A}{2} + \sin^{2} (90^{\circ} - \frac{A}{2})$.Using the identity $\sin(90^{\circ} - 	heta) = \cos 	heta$,we get:$\sin^{2} \frac{A}{2} + \cos^{2} \frac{A}{2}$.Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,the value is $1$.

If $A$ and $B$ are supplementary angles,then $\sin^{2} \frac{A}{2} + \sin^{2} \frac{B}{2} =$

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