Find the derivative of $\displaystyle h(x)$ where $\displaystyle h(x)=\frac{1}{f(x)}$ and $\displaystyle f(x)=tanx+1$, $\displaystyle 0<x<\frac{\pi}{2}$

Hence show that $\displaystyle h(x)$ is a one-one function

I have found the derivative of h(x) to be $\displaystyle -\frac{sec^2x}{(tanx+1)^2}$ but am confused on how i could use this to show $\displaystyle h(x)$ is one-one