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Question
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
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Solution
Since, A and B are complementary angles, A + B = 90°
cot B + cos B
= cot (90° – A) + cos (90° – A)
= tan A + sin A
= `sinA/cosA + sinA`
= `(sinA + sinAcosA)/cosA`
= `(sinA(1 + cosA))/cosA`
= sec A sin A (1 + cos A)
= sec A sin (90° – B) [1 + cos (90° – B)]
= sec A cos B (1 + sin B)
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