Question
Prove that:
$\frac{\sin\text{x}+\sin2\text{x}}{1+\cos\text{x}+\cos2\text{x}}=\tan\text{x}$
$\frac{\sin\text{x}+\sin2\text{x}}{1+\cos\text{x}+\cos2\text{x}}=\tan\text{x}$
✓
Answer
$\text{LHS}=\frac{\sin\text{x}+\sin2\text{x}}{1+\cos\text{x}+\cos2\text{x}}$
$=\frac{\sin\text{x}+2\sin\text{x},\cos\text{x}}{\cos\text{x}+(1+\cos2\text{x})}$
$=\frac{\sin\text{x}(1+2\cos\text{x})}{\cos\text{x}+2\cos\text{x}}$
$=\frac{\sin\text{x}(1+2\cos\text{x})}{\cos\text{x}(1+2\cos\text{x})}$
$=\frac{\sin\text{x}}{\cos\text{x}}$
$=\tan\text{x}=\text{RHS}$
$=\frac{\sin\text{x}+2\sin\text{x},\cos\text{x}}{\cos\text{x}+(1+\cos2\text{x})}$
$=\frac{\sin\text{x}(1+2\cos\text{x})}{\cos\text{x}+2\cos\text{x}}$
$=\frac{\sin\text{x}(1+2\cos\text{x})}{\cos\text{x}(1+2\cos\text{x})}$
$=\frac{\sin\text{x}}{\cos\text{x}}$
$=\tan\text{x}=\text{RHS}$
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