Simplify the trigonometric expression: (1 - cos 2x)/(sin 2x) * (sin 2x)/(1 + cos 2x)

We need to simplify:

$$\left(\frac{1-\cos 2x}{\sin 2x}\right) \left(\frac{\sin 2x}{1+\cos 2x}\right)$$

Recall the following identities:

• $1 - \cos 2x = 2\sin^2 x$
• $1 + \cos 2x = 2\cos^2 x$
• $\sin 2x = 2\sin x \cos x$

For the first fraction $\frac{1-\cos 2x}{\sin 2x}$:

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

Cancel out $2\sin x$:

$$\frac{\sin x}{\cos x} = \tan x$$

For the second fraction $\frac{\sin 2x}{1+\cos 2x}$:

$$\frac{\sin 2x}{1+\cos 2x} = \frac{2\sin x \cos x}{2\cos^2 x}$$

Cancel out $2\cos x$:

$$\frac{\sin x}{\cos x} = \tan x$$

Now multiply the simplified first and second parts:

$$(\tan x)(\tan x) = \tan^2 x$$

The final answer is $\tan^2 x$.