To prove the given trigonometric identity:

\[ \frac{2 + \sin^2 \alpha}{1 - \sin^2 \alpha} = 3 \tan^2 \alpha + 2 \]

we will use the Pythagorean identity and the definition of the tangent function.

First, recall the Pythagorean identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
which can be rearranged to:
\[ \cos^2 \alpha = 1 - \sin^2 \alpha \]

Also, recall the definition of the tangent function:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
and thus:
\[ \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \]

Now, substitute \( \cos^2 \alpha = 1 - \sin^2 \alpha \) into the expression for \( \tan^2 \alpha \):
\[ \tan^2 \alpha = \frac{\sin^2 \alpha}{1 - \sin^2 \alpha} \]

Next, rewrite the left-hand side of the given identity using the Pythagorean identity:
\[ \frac{2 + \sin^2 \alpha}{1 - \sin^2 \alpha} \]

Since \( 1 - \sin^2 \alpha = \cos^2 \alpha \), we can rewrite the denominator:
\[ \frac{2 + \sin^2 \alpha}{\cos^2 \alpha} \]

Now, separate the fraction:
\[ \frac{2}{\cos^2 \alpha} + \frac{\sin^2 \alpha}{\cos^2 \alpha} \]

Recognize that:
\[ \frac{2}{\cos^2 \alpha} = 2 \sec^2 \alpha \]
and
\[ \frac{\sin^2 \alpha}{\cos^2 \alpha} = \tan^2 \alpha \]

Thus, the expression becomes:
\[ 2 \sec^2 \alpha + \tan^2 \alpha \]

Recall the identity for \( \sec^2 \alpha \):
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \]

Substitute \( \sec^2 \alpha \) into the expression:
\[ 2 (1 + \tan^2 \alpha) + \tan^2 \alpha \]
\[ = 2 + 2 \tan^2 \alpha + \tan^2 \alpha \]
\[ = 2 + 3 \tan^2 \alpha \]

Therefore, we have shown that:
\[ \frac{2 + \sin^2 \alpha}{1 - \sin^2 \alpha} = 3 \tan^2 \alpha + 2 \]

This completes the proof.