# In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$

If in a triangle ABC, $$b+c=3a$$, then $$\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$$ is equal to ?

My reference gives the solution $$2$$, but I have no clue of where to start ?

My Attempt $$\cot\dfrac{B}{2}.\cot\dfrac{C}{2}=\frac{\cos\frac{B}{2}.\cos\frac{C}{2}}{\sin\frac{B}{2}.\sin\frac{C}{2}}=\frac{\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})}{\cos(\frac{A-B}{2})-\cos(\frac{A+B}{2})}$$

Hint:

$$b+c=3a\implies\sin B+\sin C=3\sin A$$

$$\iff2\sin\dfrac{B+C}2\cos\dfrac{B-C}2=6\sin\dfrac A2\cos\dfrac A2$$

Now use $$\dfrac{B+C}2=\dfrac\pi2-\dfrac A2,\cos\dfrac{B+C}2=?$$

As $$0

$$\implies\cos\dfrac{B-C}2=3\sin\dfrac A2=3\cos\dfrac{B+C}2$$

In the standard notation we obtain: $$\sin\beta+\sin\gamma=3\sin(\beta+\gamma)$$ or $$2\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}=6\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}$$ or $$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}+\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=3\left(\cos\frac{\beta}{2}\cos\frac{\gamma}{2}-\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\right)$$ or $$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}=2\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$$ or $$\cot\frac{\beta}{2}\cot\frac{\gamma}{2}=2.$$ Also, we can use the following way: $$\cot\frac{\beta}{2}\cot\frac{\gamma}{2}=\frac{a+b+c}{b+c-a}=\frac{4a}{2a}=2.$$