Question
If A and B are complementary angles, prove that:
cotB + cosB = secA cosB (1 + sinB)
cotB + cosB = secA cosB (1 + sinB)
✓
Answer
Since, A and B are complementary angles, A + B = 90°
cotB + cosB
= cot(90° - A) + cos(90° - A)
= tanA + sinA
$=\frac{\sin A}{\cos A}+\sin A$
$=\frac{\sin A+\sin A \cos A}{\cos A}$
$=\frac{\sin A(1+\cos A)}{\cos A}$
= secA sinA (1 + cosA)
= secA sin(90° - B)(1 + cos(90° - B))
= secA cosB(1 + sinB)
cotB + cosB
= cot(90° - A) + cos(90° - A)
= tanA + sinA
$=\frac{\sin A}{\cos A}+\sin A$
$=\frac{\sin A+\sin A \cos A}{\cos A}$
$=\frac{\sin A(1+\cos A)}{\cos A}$
= secA sinA (1 + cosA)
= secA sin(90° - B)(1 + cos(90° - B))
= secA cosB(1 + sinB)
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