Question
Prove that $\frac{\sin\text{A}-2\sin^3\text{A}}{\big(2\cos^2\text{A}-\cos\text{A}\big)}=\tan\text{A}.$
✓
Answer
$\text{LHS}=\frac{\big(\sin\text{A}-2\sin^2\text{A}\big)}{\Big(2\cos^2\text{A}-\cos\text{A}\big)}$
$=\frac{\sin\text{A}\big(1-2\sin^2\text{A}\big)}{\cos\text{A}\big(2\cos^2\text{A}-1\big)}$
$=\tan\text{A}\Bigg\{\frac{\big(\sin^2\text{A}+\cos^2\text{A}-2\sin^2\text{A}\big)}{2\cos^2\text{A}-\sin^2\text{A}-\cos^2\text{A}}\Bigg\}$ $\big[\because\sin^2\text{A}+\cos^2\text{A}=1\big]$
$=\tan\text{A}\Bigg\{\frac{\big(\cos^2\text{A}-\sin^2\text{A}\big)}{\big(\cos^2\text{A}-\sin^2\text{A}\big)}\Bigg\}$
$=\tan\text{A}$
$=\text{RHS}$
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