The value of \(\frac{1-\tan^2 45°}{1+\tan^2 45°}\) is:
Answer: D
We know that \(\tan 45° = 1\).
Substitute this value into the expression:
\(\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\).
Alternatively, this is the formula for \(\cos(2\theta)\), so it equals \(\cos(2 \times 45°) = \cos 90° = 0\).