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 ?

$$\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}$$
$$\dfrac{a^2-b^2}{a^2+b^2} = \frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}$$
$$\dfrac{a^2-b^2+a^2+b^2}{a^2-b^2-a^2-b^2} = \frac{\sin A\cos B-\cos A\sin B+\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B-\sin A\cos B-\cos A\sin B}$$
$$\dfrac{a^2}{b^2} = \frac{\sin A\cos B}{\cos A\sin B}$$
$$\dfrac{a^2}{b^2} = \frac{ak.\cos B}{\cos A.bk}$$
Because,
$$\text{sine rule}\implies\frac{\sin A}a=\frac{\sin B}b=k$$
$$\dfrac{a}{b} = \frac{\cos B}{\cos A}$$
$$a\cos A = b\cos B$$
$$\sin A\cos A=\sin B\cos B$$
$$\sin 2A=\sin 2B$$
Or, A=B.
So, the triangle is an isoceles.
