What could it be?

Use $a=2R\sin A, b=2R\sin B$ and then cancel $2R$ from both sides followed by using $\color{#20A900}{\sin A\cos B-\cos A\sin B=\sin (A-B)}$ in LHS and $\color{#D61F06}{\sin^2 A-\sin^2 B=\sin(A-B)\sin(A+B)}$ in RHS.

$\sin(A-B)=\sin(A-B)\sin(A+B)$ $\implies\sin(A-B)(\sin(A+B)-1)=0$ $\implies \sin(A-B)=0\text{ OR } \sin(A+B)=1$

First condition gives A=B (i.e Isosceles triangle) and second condition gives A+B= $\dfrac{\pi}{2}$ (i.e Right-angled triangle). Thus the correct option is: Isoceles or Right - angled

Using the sine rule,
$\dfrac{a}{\sin(A)} = \dfrac{b}{\sin(B)} = \dfrac{c}{\sin(C)} = 2R$ where R is the circumradius.

And the cosine rule,
$\cos(A) = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$
$a \left( \dfrac{a^{2}+c^{2}-b^{2}}{2ac} \right) - b\left( \dfrac{b^{2} + c^{2} - a^{2}}{2bc}\right) = \dfrac{a^{2}}{2R} - \dfrac{b^{2}}{2R}$

$\dfrac{2(a^{2}-b^{2})}{c} = \dfrac{a^{2}-b^{2}}{R}$
$\therefore a^{2}-b^{2} = 0 \to a = b$
Or
$\dfrac{c}{2R} = 1 \to \sin(C) = 1 \to C = 90^\circ$
Thus the triangle is either isoceles or right angled.