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Question
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Solution
\[LHS = \frac{\cos2x}{1 + \sin2x}\]
\[= \frac{\cos^2 x - \sin^2 x}{\sin^2 x + \cos^2 x + 2\text{ sin } x \times \text{ cos } x} \left[ \because \cos2x = \cos^2 x - \sin^2 x \text{ and } \sin^2 x + \cos^2 x = 1 \right]\]
\[ = \frac{\left( \text{ cos } x - \text{ sin } x \right)\left( \text{ cos } x + \text{ sin } x \right)}{\left( \text{ cos } x + \text{ sin } x \right)^2} \left[ \because a^2 - b^2 = \left( a + b \right)\left( a - b \right) \right]\]
\[ = \frac{\text{ cos } x - \text{ sin } x}{\text{ cos } x + \text{ sin } x}\]
On dividing the numerator and denominator by cos x, we get
\[= \frac{1 - \text{ tan } x}{1 + \text{ tan } x}\]
\[ = \tan\left( \frac{\pi}{4} - x \right) = RHS\]
\[\text{ Hence proved } .\]
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