Question

$\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A\sin B}=2$.

Answer 1

On simplification, we have to prove that $\frac{\sin A\cos A+\sin B\cos B+\sin C\cos C}{\sin A\sin B\sin C}=2$.
Multiplying both sides by $2$, we get
$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$.