$\displaystyle g(x)=tan^{-1} x+ tan^{-1} (\frac{1}{x})$. Find $\displaystyle g'(x)$ and hence deduce that $\displaystyle tan^{-1} x + tan^{-1} (\frac{1}{x}) = \frac{\pi}{2}$
$\displaystyle
g(x)=tan^{-1} x+ tan^{-1} (\frac{1}{x})
$
$\displaystyle
g'(x)=\frac{1}{1+x^2}-\frac{1}{1+\frac{1}{x^2}}(\frac{-1}{x^2})=0
$
Since g'(x)=0,therefore g(x) must be a constant.
$\displaystyle
g(x)=g(1)=tan^{-1}1+tan^{-1}1=\frac{\pi}{2}
$
Now,you should have mentioned that $\displaystyle x>0$.
Otherwise if $\displaystyle x<0$ then $\displaystyle g(x)=-\frac{\pi}{2}$