Answer
$\tan^{-1} x+\tan^{-1} (\dfrac{1}{x})=\dfrac{\pi}{2}$
Work Step by Step
Need to prove $\tan^{-1} x+\tan^{-1} \dfrac{1}{x}=\dfrac{\pi}{2}$
Consider $\tan^{-1} x=p$ and $\tan^{-1} \dfrac{1}{x}=q$
This gives: $\tan p=x$ and $\tan q=\dfrac{1}{x}$
$\tan (p+q)=\dfrac{\tan p+\tan q}{1-\tan p\tan q}$
or, $\tan (p+q)=\dfrac{x+\dfrac{1}{x}}{1-x(\dfrac{1}{x})}$
or, $(p+q)=\tan^{-1}[\infty]$
Thus, $\tan^{-1} x+\tan^{-1} (\dfrac{1}{x})=\dfrac{\pi}{2}$
Hence, the result has been proved.
