sin ^{2} 17.5^{circ} + sin ^{2} 72.5^{circ} i | vedclass

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(B) We have,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$ Since $\sin(90^{\circ} - 	heta) = \cos 	heta$,we can write $\sin(72.5^{\circ}) = \sin(

Vedclass · Accepted Answer

$	an ^{2} 45^{\circ}$

Vedclass · Answer

(B) We have,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$ Since $\sin(90^{\circ} - 	heta) = \cos 	heta$,we can write $\sin(72.5^{\circ}) = \sin(90^{\circ} - 17.5^{\circ}) = \cos(17.5^{\circ})$. Therefore,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ} = \sin ^{2} 17.5^{\circ} + \cos ^{2} 17.5^{\circ}$. Using the identity $\sin ^{2} 	heta + \cos ^{2} 	heta = 1$,we get $1$. Since $	an 45^{\circ} = 1$,then $	an ^{2} 45^{\circ} = 1^{2} = 1$. Thus,the expression is equal to $	an ^{2} 45^{\circ}$.

$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$ is equal to

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