Question
If A and B are complementary angles, prove that:
$\cot A \cot B-\sin A \cos B-\cos A \sin B=0$
$\cot A \cot B-\sin A \cos B-\cos A \sin B=0$
✓
Answer
$\text { Since, } A \text { and } B \text { are complementary angles, } A+B=90^{\circ}$
$\cot A \cot B-\sin A \cos B-\cos A \sin B$
$=\cot A \cot \left(90^{\circ}-A\right)-\sin A \cos \left(90^{\circ}-A\right)-\cos A \sin \left(90^{\circ}-A\right)$
$=\cot A \tan A-\sin A \sin A-\cos A \cos A$
$=1-\left(\sin ^2 A+\cos ^2 A\right)$
$=1-1$
$=0$
$\cot A \cot B-\sin A \cos B-\cos A \sin B$
$=\cot A \cot \left(90^{\circ}-A\right)-\sin A \cos \left(90^{\circ}-A\right)-\cos A \sin \left(90^{\circ}-A\right)$
$=\cot A \tan A-\sin A \sin A-\cos A \cos A$
$=1-\left(\sin ^2 A+\cos ^2 A\right)$
$=1-1$
$=0$
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