Question
Without using trignometric tables, prove that:
$\tan^266^\circ+\cot^224^\circ=0$
$\tan^266^\circ+\cot^224^\circ=0$
✓
Answer
$\text{L.H.S.}=\tan^266^\circ+\cot^224^\circ$
$=\tan^2\big(90^\circ-24^\circ\big)-\cot^224^\circ$
$=\cot^224^\circ-\cot^224^\circ$
$=0$
$=\text{R.H.S}$
$=\tan^2\big(90^\circ-24^\circ\big)-\cot^224^\circ$
$=\cot^224^\circ-\cot^224^\circ$
$=0$
$=\text{R.H.S}$
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