Question

Let \( f(x) = \cos^{-1} \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) \). Then \( f\left( \frac{\pi}{2} \right) \) is equal to:

• A. -1
• B. 2
• C. 1
• D. \( \frac{\sqrt{3}}{2} \)
• E. \( \sqrt{3} \)

Hint:

We use the identity \( \frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos(2x) \) to simplify the expression.

Answer 1

The Correct Option is B

We are given the function \( f(x) = \cos^{-1} \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) \). 
Using the trigonometric identity \( \frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos(2x) \), we can rewrite the function as: \[ f(x) = \cos^{-1}(\cos(2x)). \] Now, we need to evaluate \( f\left( \frac{\pi}{2} \right) \): \[ f\left( \frac{\pi}{2} \right) = \cos^{-1}(\cos(\pi)) = \cos^{-1}(-1) = 2. \] Thus, \( f\left( \frac{\pi}{2} \right) = 2 \).