Answer
$\frac{1}{sec~x+1}-\frac{1}{sec~x-1}=--2~cot^2x$
Work Step by Step
$\frac{1}{sec~x+1}-\frac{1}{sec~x-1}=\frac{1(sec~x-1)}{(sec~x+1)(sec~x-1)}-\frac{1(sec~x+1)}{(sec~x-1)(sec~x+1)}=\frac{sec~x-1}{sec^2x-1}-\frac{sec~x+1}{sec^2x-1}=\frac{sec~x-1-sec~x-1}{sec^2x-1}=\frac{-2}{tan^2x}=-2~cot^2x$
