Question
Prove that:
$\sin(70^\circ+\theta)-\cos(20^\circ-\theta)=0$
$\sin(70^\circ+\theta)-\cos(20^\circ-\theta)=0$
✓
Answer
$\text{L.H.S.}=\sin(70^\circ+\theta)-\cos(20^\circ-\theta)$
$=\sin\big\{90^\circ-(20^\circ-\theta)\big\}-\cos(20^\circ-\theta)$
$=\cos(20^\circ-\theta)-\cos(20^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
$=\sin\big\{90^\circ-(20^\circ-\theta)\big\}-\cos(20^\circ-\theta)$
$=\cos(20^\circ-\theta)-\cos(20^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
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