<s>[INST] Which triangle follows the condition of $2\tan \frac{A}{2}=\frac{\sin A}{\sin B\sin C}$? The problem is as follows:
Let a triangle $A$, $B$ and $C$, where it follows this condition.
$2\tan \frac{A}{2}=\frac{\sin A}{\sin B\sin C}$
On which of the following triangles such condition is met?
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{Isosceles triangle}\\
2.&\textrm{Right triangle}\\
3.&\textrm{Acute triangle}\\
4.&\textrm{Right isosceles triangle}\\
\end{array}$
This problem is somewhat complicated to my understanding. As I don't know exactly where to begin with.
$2\csc A - 2\cot A= \frac{\sin A}{\sin B \sin C}$
Then:
$2\frac{1}{\sin A}-2\frac{\cos A}{\sin A} =\frac{\sin A}{\sin B \sin C}$
But that's where I could only reach because from further expanding the earlier equation it doesn't seem to get useful information.
Can someone help me here, on how to arrive to the conclusion of which of those triangles mentioned would match that peculiar rule?.
It would help a lot an answer which could explain the most details as possible so to avoid getting confused with the steps.
 [/INST] You have: $2\tan(\frac{A}{2}) = \dfrac{\sin A}{\sin B\sin C}\implies 2\dfrac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}= \dfrac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin B\sin C}\implies \sin B\sin C = \cos^2(\frac{A}{2})= \dfrac{1+\cos A}{2}\implies 1+\cos A = 2\sin B\sin C= \cos(B-C)-\cos(B+C)= \cos(B-C)+\cos A\implies \cos(B-C) = 1\implies B-C=0\implies B = C$. Thus triangle $ABC$ is isosceles.
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