Question

Prove : $\sin$ $\theta \sin ({90}^{\circ }-\theta )-\cos \theta \cos ({90}^{\circ }-\theta )=0.$

Answer 1

We know;

$\sin ({90}^{\circ }-\theta )=\cos \theta and\cos ({90}^{\circ }-\theta )=\sin \theta$

L.H.S$=\sin \theta \sin ({90}^{\circ }-\theta )-\cos \theta \cos ({90}^{\circ }-\theta )$

$=\sin \theta \cos \theta -\cos \theta \sin \theta$

$=0$=R.H.S

$\because L.H.S=R.H.S$

$\sin \theta \sin ({90}^{\circ }-\theta )-\cos \theta \cos ({90}^{\circ }-\theta )=0$

Hence proved