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