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Prove the following:
$ \sin \theta \sin \left(90^{\circ}-\theta\right)-\cos \theta \cos \left(90^{\circ}-\theta\right)=0 $
To do:
We have to prove that \( \sin \theta \sin \left(90^{\circ}-\theta\right)-\cos \theta \cos \left(90^{\circ}-\theta\right)=0 \).
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
$cos\ (90^{\circ}- \theta) = sin\ \theta$
Therefore,
$\sin \theta \sin \left(90^{\circ}-\theta\right)-\cos \theta \cos \left(90^{\circ}-\theta\right)=\sin \theta \cos \theta-\cos \theta \sin \theta$
$=\sin \theta \cos \theta-\sin \theta \cos \theta$
$=0$
Hence proved.
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