Prove that `(1 - tan^(2) theta)/(1 + tan^(2) theta) = cos^(2) - sin^(2) theta`

Answer

To prove the equation

\[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos^2 \theta - \sin^2 \theta, \]

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

### Step 1: Substitute \(\tan \theta\)

Recall that

\[ \tan \theta = \frac{\sin \theta}{\cos \theta}. \]

Thus, we can express \(\tan^2 \theta\) as:

\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}. \]

Now, substituting this into the left-hand side gives us:

\[ \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}}. \]

### Step 2: Simplify the Expression

To simplify, we can multiply both the numerator and the denominator by \(\cos^2 \theta\):

\[ \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta}. \]

### Step 3: Use the Pythagorean Identity

We know from the Pythagorean identity that:

\[ \sin^2 \theta + \cos^2 \theta = 1. \]

Thus, the denominator simplifies to:

\[ \cos^2 \theta + \sin^2 \theta = 1. \]

### Step 4: Final Simplification

Now, substituting this back into our expression gives:

\[ \frac{\cos^2 \theta - \sin^2 \theta}{1} = \cos^2 \theta - \sin^2 \theta. \]

### Conclusion

We have shown that:

\[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos^2 \theta - \sin^2 \theta. \]

Thus, the equation is proved.

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