Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.7 Derivatives And Integrals Involving Inverse Trigonometric Functions - Exercises Set 6.7 - Page 472: 89

See the proof below.

Let $tan^{-1}x= α$ and $tan^{-1}y=β$. Then x= tan α and y= tan β. tan(α+β) =$ \frac{tan α+tan β}{1-tanαtanβ}=\frac{x+y}{1-xy}$ Which gives α+β = $tan^{-1}\frac{x+y}{1-xy}$ Or $tan^{-1}x+tan^{-1}y$=$tan^{-1}(\frac{x+y}{1-xy})$