<s>[INST] Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$, 
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$
Also we know that all the expressions are different from zero and defined. Including the expressions we received during the solution. 
Tried to play with it, didn't seem to work for me.
 [/INST] Using the formulas of half-angle substitution, also called Weierstrass substitution formulas,
$$\cos(u)=\dfrac{1-t^2}{1+t^2} , \ \sin(u)=\dfrac{2t}{1+t^2} \ \ \ \text{where} \ \ \ t:=\tan(\tfrac{u}{2}),$$
one has to prove, setting $a=\tan(\alpha/2)$ and $b=\tan(\beta/2)$, that :
$$2\dfrac{2a}{1+a^2}+2\dfrac{2b}{1+b^2}=3 \dfrac{2a}{1+a^2}\dfrac{1-b^2}{1+b^2} +3 \dfrac{2b}{1+b^2}\dfrac{1-a^2}{1+a^2} \implies ab=\dfrac15$$
(we have used $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$).
The first expression, once we have multiplied both sides by $(1+a^2)(1+b^2)$ and cancelled by 2, gives:
$$2a+2ab^2+2b+2ba^2 = 3a-3a^2b+3b-3a^2b$$
$$\iff 5ab(a+b)=a+b$$
giving indeed, under the supplementary assumption that $a+b \neq 0$ 

$$ab=\tfrac15.$$

Remark: the overall advantage of formulas (*) is that they convert trigonometrical issues into fully algebraic formulations.
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