Question
Without using trignometric tables, prove that:
$\tan71^\circ-\cot19^\circ=0$

Answer

$\text{L.H.S.}=\tan71^\circ-\cot19^\circ$
$=\tan\big(90^\circ-19^\circ\big)-\cot19^\circ$
$=\cot19^\circ-\cot19^\circ$
$=0$
$=\text{R.H.S}$

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