Question
Prove that:
$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec }31^\circ=2$
$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec }31^\circ=2$
✓
Answer
$\text{L.H.S}=\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec }31^\circ$
$=\frac{\cos80^\circ}{\cos(90^\circ-10^\circ)}+\sin(90^\circ-59^\circ)\text{cosec }31^\circ$
$=\frac{\cos80^\circ}{\cos80^\circ}+\sin31^\circ\text{cosec }31^\circ$
$=1+\sin31^\circ\times\frac{1}{\sin31^\circ}$
$=1+1$
$=2$
$=\text{R.H.S.}$
$=\frac{\cos80^\circ}{\cos(90^\circ-10^\circ)}+\sin(90^\circ-59^\circ)\text{cosec }31^\circ$
$=\frac{\cos80^\circ}{\cos80^\circ}+\sin31^\circ\text{cosec }31^\circ$
$=1+\sin31^\circ\times\frac{1}{\sin31^\circ}$
$=1+1$
$=2$
$=\text{R.H.S.}$
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