Answer
$\tan^{-1}(\dfrac{x+y}{1-xy})=\tan^{-1} x+\tan^{-1} y$
Work Step by Step
Need to prove $\tan^{-1}(\dfrac{x+y}{1-xy})=\tan^{-1} x+\tan^{-1} y$
Consider $\tan^{-1} x=p$ and $\tan^{-1} y=q$
This gives: $\tan p=x$ and $\tan q=y$
$\tan (p+q)=\dfrac{\tan p+\tan q}{1-\tan p\tan q}$
or, $\tan (p+q)=\dfrac{x+y}{1-xy}$
or, $(p+q)=\tan^{-1}[\dfrac{x+y}{1-xy}]$
Thus, $\tan^{-1}(\dfrac{x+y}{1-xy})=\tan^{-1} x+\tan^{-1} y$
Hence, the result has been proved.
