Question

What is \( \tan \left( \frac{1}{2} \left( \tan^{-1} x + \tan^{-1} \frac{1}{x} \right) \right) \)?

• A. 1
• B. \( \sqrt{3} \)
• C. 0
• D. \( \infty \)

Hint:

When dealing with inverse tangent functions, use the identity for \( \tan^{-1} a + \tan^{-1} b \) to simplify the expression.

Answer 1

The Correct Option is A

We are asked to find: \[ \tan \left( \frac{1}{2} \left( \tan^{-1} x + \tan^{-1} \frac{1}{x} \right) \right) \] We know the formula for \( \tan^{-1} a + \tan^{-1} b \) is: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \] provided that \( ab& Lt;1 \). For our case: \[ \tan^{-1} x + \tan^{-1} \frac{1}{x} = \tan^{-1} \left( \frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}} \right) \] \[ = \tan^{-1} \left( \frac{x + \frac{1}{x}}{0} \right) = \tan^{-1} (\infty) \] The value of \( \tan^{-1} (\infty) \) is \( \frac{\pi}{2} \). Thus: \[ \tan \left( \frac{1}{2} \cdot \frac{\pi}{2} \right) = \tan \left( \frac{\pi}{4} \right) = 1 \] Thus, the correct answer is \( 1 \).