Question
Prove that:
$\tan(55^\circ-\theta)-\cot(35^\circ+\theta)=0$
$\tan(55^\circ-\theta)-\cot(35^\circ+\theta)=0$
✓
Answer
$\text{L.H.S.}=\tan(55^\circ-\theta)-\cot(35^\circ+\theta)$
$=\tan\big\{90^\circ-(35^\circ-\theta)\big\}-\cot(35^\circ-\theta)$
$=\cot(35^\circ+\theta)-\cot(35^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
$=\tan\big\{90^\circ-(35^\circ-\theta)\big\}-\cot(35^\circ-\theta)$
$=\cot(35^\circ+\theta)-\cot(35^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
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