Prove that (sin A + cos A)/(sin A – cos A) + (sin A

Q) Prove that $\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{2 sin2 A - 1}$

Ans: Let’s start solving from LHS:

$\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A}$

= $\frac{(sin A + cos A)(sin A + cos A) + (sin A - cos A)(sin A - cos A)}{(sin A - cos A)(sin A + cos A)}$

= $\frac{(sin ^2 A + cos ^2 A + 2 sin A cos A) + (sin ^2 A + cos ^2 A - 2 sin A cos A)}{(sin^2 A - cos^2 A)}$

= $\frac{(sin ^2 A + cos ^2 A + 2 sin A cos A + sin ^2 A + cos ^2 A - 2 sin A cos A)}{(sin^2 A - cos^2 A)}$

= $\frac{(sin ^2 A + cos ^2 A + \cancel{2 sin A cos A} + sin ^2 A + cos ^2 A - \cancel{2 sin A cos A})}{(sin^2 A - cos^2 A)}$

= $\frac{2 (sin ^2 A + cos ^2 A)}{(sin^2 A - cos ^2 A)}$

∵ sin² A + cos² A = 1, ∴ cos² A = 1 – sin² A

∴ LHS = $\frac{2 (1)}{(sin^2 A - (1 - sin ^2 A)}$

= $\frac{2 (1)}{(sin^2 A - 1 + sin ^2 A)}$

= $\frac{2}{2 sin^2 A - 1}$

= RHS…. Hence Proved !

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