Question

Prove that \[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1 - xy} \right), \] provided \( xy<1 \).

Hint:

Use the tangent addition formula: \[ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] to prove identities involving inverse tangents.

Answer 1

Let: \[ \alpha = \tan^{-1} x, \quad \beta = \tan^{-1} y. \] Then, \[ \tan \alpha = x, \quad \tan \beta = y. \] Using the tangent addition formula: \[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{x + y}{1 - xy}. \] Since \(\alpha = \tan^{-1} x\) and \(\beta = \tan^{-1} y\), it follows that: \[ \alpha + \beta = \tan^{-1} \left( \frac{x + y}{1 - xy} \right), \] provided the expression is defined (i.e., \(xy<1\)).