Question
Without using trigomentric tables, prove that:
$\big(\sin72^\circ+\cos18^\circ\big)\big(\sin72^\circ-\cos18^\circ\big)=0$

Answer

$\text{L.H.S}=\big(\sin72^\circ+\cos18^\circ\big)\big(\sin72^\circ-\cos18^\circ\big)$
$=\big(\sin72^\circ+\cos18^\circ\big)\big[\cos\big(90^\circ-72^\circ\big)-\cos18^\circ\big]$
$=\big(\sin72^\circ+\cos18^\circ\big)\big(\cos18^\circ-\cos18^\circ\big)$
$=\big(\sin72^\circ+\cos18^\circ\big)(0)$
$=0$
$=\text{R.H.S.}$

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