how to verify $\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x))}$? How would I verifty the following trig identity?

$$
\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}
$$

I am not sure how to start.
 A: $$\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\frac{\sin(x)\cos(x)}{\cos^2 x}}{\frac{\cos^2(x)-\sin^2(x)}{\cos^2x}}=$$
$$=\frac{\tan(x)}{1-\tan^2(x)}$$
A: $$\frac{2\tan(x)}{1-\tan^2(x)}=\tan(2x),$$
$$\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\sin(2x)}{\cos(2x)}.$$
A: Hint:
$$\frac{\tan(x)}{1-\tan^2(x)}=\frac{\dfrac{\sin(x)}{\cos(x)}}{1-\dfrac{\sin^2(x)}{\cos^2(x)}}$$ and you are at one millimeter from the solution.
A: $$\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}\Longleftrightarrow$$
$$\frac{\sec^2(x)}{\sec^2(x)}\cdot\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}\Longleftrightarrow$$
$$\frac{\tan(x)}{\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)}=\frac{\tan(x)}{1-\tan^2(x)}$$
Now, use $\sec(x)=\frac{1}{\cos(x)}$:
$$\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)=\cos^2(x)\sec^2(x)-\sin^2(x)\sec^2(x)=$$
$$\left(\frac{1}{\cos(x)}\right)^2\cos^2(x)-\left(\frac{1}{\cos(x)}\right)^2\sin^2(x)=1-\frac{\sin^2(x)}{\cos^2(x)}=1-\tan^2(x)$$
So, we can conclude:
$$\frac{\tan(x)}{\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)}=\frac{\tan(x)}{1-\tan^2(x)}$$
