We are asked to prove the trigonometric identity: $1 - 2\cos^2{t} = \frac{\tan^2{t} - 1}{\tan^2{t} + 1}$.

1. Problem Description

We are asked to prove the trigonometric identity:

$1 - 2\cos^2{t} = \frac{\tan^2{t} - 1}{\tan^2{t} + 1}$.

2. Solution Steps

We start with the right-hand side of the equation and try to simplify it to match the left-hand side.

We know that $\tan{t} = \frac{\sin{t}}{\cos{t}}$, so $\tan^2{t} = \frac{\sin^2{t}}{\cos^2{t}}$.

Substituting this into the right-hand side of the equation gives us:

$\frac{\tan^2{t} - 1}{\tan^2{t} + 1} = \frac{\frac{\sin^2{t}}{\cos^2{t}} - 1}{\frac{\sin^2{t}}{\cos^2{t}} + 1}$

To simplify this expression, we can multiply both the numerator and the denominator by $\cos^2{t}$:

$\frac{\frac{\sin^2{t}}{\cos^2{t}} - 1}{\frac{\sin^2{t}}{\cos^2{t}} + 1} = \frac{(\frac{\sin^2{t}}{\cos^2{t}} - 1) \cdot \cos^2{t}}{(\frac{\sin^2{t}}{\cos^2{t}} + 1) \cdot \cos^2{t}} = \frac{\sin^2{t} - \cos^2{t}}{\sin^2{t} + \cos^2{t}}$

We know the Pythagorean identity:

$\sin^2{t} + \cos^2{t} = 1$

Therefore,

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

Also, we have the identity $\sin^2{t} = 1 - \cos^2{t}$.

Substituting this in gives us:

$\sin^2{t} - \cos^2{t} = (1 - \cos^2{t}) - \cos^2{t} = 1 - 2\cos^2{t}$

This matches the left-hand side of the equation, so we have shown that:

$1 - 2\cos^2{t} = \frac{\tan^2{t} - 1}{\tan^2{t} + 1}$

3. Final Answer

$1 - 2\cos^2{t} = \frac{\tan^2{t} - 1}{\tan^2{t} + 1}$