algebra precalculus - If $frac{a}{sin{A}}=frac{b}{cos{A}}$, show that $sin{A}cos{A}=frac{ab}{a^2+b^2}$

I don't know how to go about solving this, I think I need to use $\sin^2\theta+\cos^2\theta=1$, but I'm not sure how to go about this.

The closest I managed to get was:

$$\frac{a}{\sin{A}}=\frac{b}{\cos{A}}$$
$$a\cos{A}-b\sin{A}=0$$
$$a^2\cos^2{A}+b^2\sin^2{A}-2ab\cos{A}\sin{A}=0$$
$$\cos{A}\sin{A}=\frac{a^2\cos^2{A}+b^2\sin^2{A}}{2ab}$$

Which seems to vaguely resemble what I need, but I'm not sure of the final steps.

Answer

Notice that
$$\sin A \cos A = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A} = \frac{\tan A}{\tan^2 A + 1}$$
It should be ease for you to continue.