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Question
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
Solution
\[LHS = \frac{1 - \cos2x + \sin2x}{1 + \cos2x + \sin2x}\]
\[= \frac{2 \sin^2 x + \sin2x}{2 \cos^2 x + \sin2x} \left[ \because 2 \sin^2 x = 1 - \cos2x and 2 \cos^2 x = 1 + \cos2x \right]\]
\[ = \frac{2 \sin^2 x + 2\text{ sin } x \text{ cos } x}{2 \cos^2 x + 2\text{ sin } x \text{ cos } x} \left( \because \sin2x = 2\text{ sin } x \text{ cos } x \right)\]
\[ = \frac{2\text{ sin } x\left( \text{ sin } x + \text{ cos } x \right)}{2\text{ cos } x\left( \text{ cos } x + \text{ sin } x \right)} \]
\[ = tan\theta = RHS\]
\[\text{ Hence proved }.\]
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`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta = (2tantheta)/(1 + tan^2theta)]`.
