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Trigonometric Ratios of Compound Angles

Prove that: ...

Question

Prove that:
$\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{2 {sin}^{2} A - 1}$

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Solution

L.H.S $= \frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A}$
$= \frac{{sin}^{2} A + {cos}^{2} A + 2 sin A cos A + {sin}^{2} A + {cos}^{2} A - 2 sin A cos A}{{sin}^{2} A - {cos}^{2} A}$
$= \frac{2 ({sin}^{2} A + {cos}^{2} A)}{{sin}^{2} A - (1 - {sin}^{2} A)}$
$= \frac{2}{1 - 2 {sin}^{2} A} =$ R.H.S
Hence proved.

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