Verified Answer :

dalendrk

We know:
\(\sin(90^o-\alpha)=\cos\alpha\)

\(\sin25^o=\sin(90^o-65^o)=\cos65^o\ therefore\ \sin^225^o=\cos^265^o\)

We have:
\(\sin^225^o+\sin^265^o=\cos^265^o+\sin^265^o=1\)

Used: \(\sin^2\alpha+\cos^2\alpha=1\)

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