If $ \dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)} $, then what type of triangle is $\triangle ABC $? In $\triangle ABC$
$
\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}
$
then what type of triangle is  $\triangle ABC $  ?

My try :
By componendo and dividendo 
$\dfrac{a^2}{b^2} = \tan A \cot B$
Not able to conclude, any help ?

You are nearly there. Just apply the sine rule in the form $$\dfrac ab=\dfrac{\sin A}{\sin B}$$to get $\sin A\cos A=\sin B\cos B$; that is, $\sin 2A=\sin 2B$ and hence either $A=B$ or $C$ ($=\pi-A-B$) =$\pi/2$.
Edit Apologies to the OP for an incorrect first answer.
