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Basic Trigonometric Identities

Prove the fol...

Question

Prove the following identity:
$\frac{1 - 2 {sin}^{2} α}{1 + sin 2 α} = \frac{1 - tan α}{1 + tan α}$

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Solution

$\frac{1 - 2 {sin}^{2} α}{1 + sin 2 α}$
$= \frac{cos 2 α}{{(sin α + cos α)}^{2}}$
$= \frac{{cos}^{2} α - {sin}^{2} α}{{(cos α + sin α)}^{2}}$
$= \frac{cos θ - sin θ}{cos θ + sin θ}$
$= \frac{1 - tan θ}{1 + tan θ}$

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