We are tasked with proving that:

$\sqrt{\frac{1+x}{1-x}} = |\cot \theta|$

where $x = \cos 2\theta$. Let's work through this step by step using trigonometric identities.

Step 1: Use the double angle identity for cosine

We know the double angle identity for cosine is:

$\cos 2\theta = 2\cos^2\theta - 1$

Thus, $x = \cos 2\theta$ can be written as:

$x = 2\cos^2\theta - 1$

Step 2: Express $\cos^2\theta$ in terms of $x$

Solving for $\cos^2\theta$ from the above equation:

$x + 1 = 2\cos^2\theta$

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

Step 3: Use the identity $\sin^2\theta + \cos^2\theta = 1$

Now, using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, we can express $\sin^2\theta$ in terms of $\cos^2\theta$:

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

Step 4: Express $\cot \theta$ in terms of $x$

Recall that $\cot \theta = \frac{\cos \theta}{\sin \theta}$. Using the expressions for $\cos^2\theta$ and $\sin^2\theta$, we have:

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

Taking the square root on both sides:

$|\cot \theta| = \sqrt{\frac{x+1}{1-x}}$

Step 5: Compare both sides

From the previous steps, we have:

$|\cot \theta| = \sqrt{\frac{x+1}{1-x}}$

This matches the required expression:

$\sqrt{\frac{1+x}{1-x}} = |\cot \theta|$

Thus, the proof is complete.

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Would you like more details on any step, or do you have any questions?

Here are 5 related questions:

1. How is the identity for $\cos 2\theta$ derived?
2. Can we use a similar approach to prove an identity involving $\tan \theta$?
3. What other identities are related to $\cot \theta$?
4. How does the expression for $\sin 2\theta$ differ from $\cos 2\theta$?
5. Can this proof be extended to other trigonometric functions?

Tip: Remember that understanding and using double-angle identities is key to simplifying many trigonometric problems.