Question

Find \( \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{3} \right) \).

• A. \( \pi \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{\pi}{3} \)

Hint:

Use the identity \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \) for simplifying the sum of inverse tangents.

Answer 1

The Correct Option is B

We are given \( \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{3} \right) \). We can use the formula for the sum of inverse tangents: \[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \] So: \[ \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{3} \right) = \tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \times \frac{1}{3}\right)} \right) \] First, simplify the numerator: \[ \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \] And the denominator: \[ 1 - \left(\frac{1}{2} \times \frac{1}{3}\right) = 1 - \frac{1}{6} = \frac{5}{6} \] So the expression becomes: \[ \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) = \tan^{-1} (1) = \frac{\pi}{4} \] Thus, the correct answer is \( \frac{\pi}{4} \).