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How would I prove the following double angle identity?

$$\frac{\sin2A - \sin A}{\cos2A + \sin A}=\tan \frac{3A}{2} . \cot\frac{A}{2}$$

Sadly I am stuck.

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This is similar to one of the tangent half-angle formulas: $\dfrac{\sin A+\sin B}{\cos A+\cos B}=\tan\dfrac{A+B}{2}$. – Michael Hardy May 5 '13 at 22:41
@Mob The half-angle identity $\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and the similar identity for $\cot(\frac{\theta}{2})$ might be useful on the right-hand side. – Stefan Smith May 5 '13 at 23:31

This is simply wrong, if there is no typo. Take $A=\frac{\pi}{3}$, $LHS=0\neq RHS=\infty$.

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RHS=$\dfrac{sin2A+sinA}{sin2A-sinA}$, so only some $A$ will let LHS=RHS. but it is 6 degree equation.are you sure you have right information?

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$tan(\frac{3A}{2}) * cot(\frac{A}{2})$

$\Rightarrow$ $\frac{sin(\frac{3A}{2}) * cos(\frac{A}{2})}{cos(\frac{3A}{2}) * sin(\frac{A}{2})}$

$\Rightarrow$ $\frac{sin(\frac{4A}{2}) + sin(\frac{2A}{2})}{sin(\frac{4A}{2}) - sin(\frac{2A}{2})}$

$\Rightarrow$ $\frac{sin(2A) + sin(A)}{sin(2A) - sin(A)}$

$\Rightarrow$ $\frac{2*cos(A) + 1}{2*cos(A) - 1}$

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