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Question
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
Solution
Since, A and B are complementary angles, A + B = 90°
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA)`
= `(sinA + sinB)/(sinA - sinB) + (cos(90^@ - A) - cos(90^@ - B))/(cos(90^@ - A) + cos(90^@ - B))`
= `(sinA + sinB)/(sinA - sinB) + (sinA - sinB)/(sinA + sinB)`
= `((sinA + sinB)^2 + (sinA - sinB)^2)/((sinA - sinB)(sinA + sinB)`
= `(sin^2A + sin^2B + 2sinAsinB + sin^2A + sin^2B - 2sinA)/(sin^2A - sin^2B`
= `2(sin^2A + sin^2B)/(sin^2A - sin^2B)`
= `2(sin^2A + sin^2(90^@ - A))/(sin^2A - sin^2(90^@ - A))`
= `2(sin^2A + cos^2B)/(sin^2A - cos^2B)`
= `2/(sin^2A - (1 - sin^2A))`
= `2/(2sin^2A - 1)`
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