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Question
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Solution
LHS = `(sin B + sec A)/sin A`
= `(sin (90 - A) + sec A)/sin A`
= `(cos A + sec A)/sin A`
= `(cos^2 A + 1)/(sin A. cos A)`
= `(2cos^2 A + sin^2 A)/(sin A. cos A)`
= 2cot A + tan A
= 2 tan B + tan A = RHS
hence proved.
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