Question

Prove that: \[ \tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{3}\right) = \frac{\pi}{4}. \]

Hint:

Use the identity: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a + b}{1 - ab}\right). \] for sum of inverse tangent functions.

Answer 1

Step 1: Use the Identity
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a + b}{1 - ab}\right), \quad {if } ab<1. \] Step 2: Apply Given Values
\[ \tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{3}\right) = \tan^{-1} \left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right). \] \[ = \tan^{-1} \left(\frac{\frac{3 + 2}{6}}{1 - \frac{1}{6}}\right). \] \[ = \tan^{-1} \left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1} (1). \] \[ = \frac{\pi}{4}. \]