Question
Prove that:
$\sin\theta\cos(90^\circ-\theta)+\sin(90^\circ-\theta)\cos\theta=1$
$\sin\theta\cos(90^\circ-\theta)+\sin(90^\circ-\theta)\cos\theta=1$
✓
Answer
$\text{L.H.S.}=\sin\theta\cos(90^\circ-\theta)+\sin(90^\circ-\theta)\cos\theta$
$=\sin\theta\sin\theta+\cos\theta\cos\theta$
$=\sin^2\theta+\cos^2\theta$
$=1$
$=\text{R.H.S.}$
Hence Proved.
$=\sin\theta\sin\theta+\cos\theta\cos\theta$
$=\sin^2\theta+\cos^2\theta$
$=1$
$=\text{R.H.S.}$
Hence Proved.
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