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Question
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
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Solution
\[LHS = \frac{\text{ sin } x + \sin2x}{1 + \text{ cos } x + \cos2x}\]
\[ = \frac{\text{ sin } x + \sin2x}{\text{ cos } x + \left( 1 + \text{ cos } 2x \right)}\]
\[= \frac{\text{ sin } x + 2\text{ sin } x \text{ cos } x}{\text{ cos } x + 2 \cos^2 x} \left[ \because \sin2x = 2\text{ sin } x \text{ cos } x \text{ and } 2 \cos^2 x = 1 + \cos2x \right]\]
\[= \frac{\text{ sin } x \left( 1 + 2\text{ cos } x \right)}{\text{ cos } x \left( 1 + 2\text{ cos } x \right)}\]
\[ = \text{ tan } x = RHS\]
\[\text{ Hence proved } .\]
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