(d) $\sin 2A + \sin 2B\, – \sin 2C$ 

$= 2\sin A\cos A + 2\cos (B + C)\sin (B – C)$

$\{ \because A + B + C = \pi ,\,\therefore \,B + C = \pi  – A,\cos (B + C) = \cos (\pi  – A),$ $\cos (B + C) = – \cos A,\,\sin (B + C) = \sin A\} $

$ = 2\cos A\,\,[\sin A – \sin (B – C)]$

$ = 2\cos A\,[\sin (B + C) – \sin (B – C)]$ 

$ = 2\cos A.2\cos B.\sin C$

$ = 4\cos A.\,\cos B.\,\sin C$.

Trick : First put $A = B = C = {60^o}$ for, these values. 

Options $(a)$ and $ (d)$ satisfies the condition, Now put $A = B = 45^\circ $ and $c = {90^o}$. 

Then only $(d)$ satisfies. Hence $(d)$ is the answer.