Prove that
`sqrt ((1- cos 2x)/( 1 + cos 2x )) = tan x, x in I or III` quad.

Answer

To prove that

\[ \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x, \quad x \text{ in I or III quadrant} \]

we will start with the left-hand side and simplify it step by step.

### Step 1: Start with the left-hand side
We begin with the expression:

\[ \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \]

### Step 2: Use the double angle identities
Recall the double angle identities for cosine:

\[ \cos 2x = 1 - 2\sin^2 x \quad \text{and} \quad \cos 2x = 2\cos^2 x - 1 \]

Using these, we can rewrite \(1 - \cos 2x\) and \(1 + \cos 2x\):

1. For \(1 - \cos 2x\):
\[ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \]

2. For \(1 + \cos 2x\):
\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \]

### Step 3: Substitute back into the expression
Now substitute these results back into the original expression:

\[ \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \sqrt{\frac{\sin^2 x}{\cos^2 x}} \]

### Step 4: Simplify the square root
The square root of a fraction can be simplified:

\[ \sqrt{\frac{\sin^2 x}{\cos^2 x}} = \frac{\sin x}{\cos x} \]

### Step 5: Recognize the tangent function
We know that:

\[ \frac{\sin x}{\cos x} = \tan x \]

### Conclusion
Thus, we have shown that:

\[ \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x \]

This completes the proof, and since \(x\) is in the first or third quadrant, \(\tan x\) is defined.