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