<s>[INST] What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ? 
 [/INST] So by Law of Sines we have $$ \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} =k (\text{say})$$
From this your equation becomes, $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{k(\sin{B} + \sin{C})}{k \sin{A}} = \frac{\sin{B}+\sin{C}}{\sin{A}} = \frac{\cos\frac{A}{2} \cos\frac{B-C}{2}}{\sin\frac{A}{2} \cos\frac{A}{2}}$$ From this you get $$\frac{\pi}{2} - \Bigl(\frac{B+C}{2}\Bigr)= \pm \frac{B-C}{2}$$ and you can do it then.
This gives $B=\frac{\pi}{2}$ or $C =\frac{\pi}{2}$, which means the triangle is right-angled. Hurray!
NOTE: The keen idea when you see problems of this type is to use the Sine rule. Do you think that this problem would have been difficult, if you had applied the sine rule as i did in the first step. Very often in Mathematics starting the solution is the difficult part. Once you figure that out, then the solutions simply flows. 
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