Question

The value of \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} \) is:

• A. \( 1 \)
• B. \( 0 \)
• C. \( -1 \)
• D. \( \infty \)

Hint:

The formula \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta \), and for \( \theta = 45^\circ \), the result is 0.

Answer 1

The Correct Option is B

Step 1: Recall the value of \( \tan 45^\circ \).
We know that \( \tan 45^\circ = 1 \). Step 2: Substitute the value of \( \tan 45^\circ \).
Substituting \( \tan 45^\circ = 1 \) into the given expression, we get: \[ \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \frac{1 - 1^2}{1 + 1^2} = \frac{0}{2} = 0. \] Step 3: Conclude the result.
Thus, the value of \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} \) is \( 0 \).