Question

Prove that $\cos \theta \sin ({90}^o-\theta )+\sin \theta \cos ({90}^o-\theta )=1$

Answer 1

L.H.S.

$\cos \theta \sin ({90}^o-\theta )+\sin \theta \cos ({90}^o-\theta )$

$=\cos \theta .\cos \theta +\sin \theta .\sin \theta$

$={\cos }^2\theta +{\sin }^2\theta =1=$ R.H.S.

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

$\cos \theta \sin ({90}^o-\theta )+\sin \theta \cos ({90}^o-\theta )=1$

Hence Proved