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To prove the identity

\[
\frac{1 + \tan^2 x}{1 - \tan^2 x} = \sec 2x,
\]

we can use trigonometric identities, particularly the tangent and secant double-angle identities.

### Step-by-Step Solution

1. **Rewrite using a known identity:**
   Recall the identity for \(\tan^2 x\):
   \[
   1 + \tan^2 x = \sec^2 x.
   \]

   So we can rewrite the left side as:
   \[
   \frac{\sec^2 x}{1 - \tan^2 x}.
   \]

2. **Rewrite \(1 - \tan^2 x\) in terms of \(\tan 2x\):**
   Another useful identity is the double-angle identity for tangent:
   \[
   \tan 2x = \frac{2 \tan x}{1 - \tan^2 x}.
   \]

3. **Rewrite the denominator using trigonometric identities:**
   If we focus on the right side, \(\sec 2x\) can also be expressed by the identity:
   \[
   \sec 2x = \frac{1}{\cos 2x}.
   \]

4.

Prove the trigonometric identity: (1 + tan^2(x)) / (1 - tan^2(x)) = sec(2x).

Explore how to prove the trigonometric identity (1 + tan^2(x)) / (1 - tan^2(x)) = sec(2x) using double-angle identities. This problem covers key trigonometric concepts, including double-angle identities for tangent and secant, and is suitable for high school students in Grades 10-12.

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