Prove that (sin A + cos A)/(sin A – cos A) + (sin A – cos A)/(sin A + cos A) = 2/(2 sin2 A – 1)

Question

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

Sapling Academy · Accepted Answer

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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