Answer
$\dfrac{\tan x+\tan {\pi/2}}{1-\tan x \tan (\pi/2)}=-\cot x$
Verified
Work Step by Step
$\dfrac{\tan x+\tan {\pi/2}}{1-\tan x \tan (\pi/2)}=\dfrac{1+\dfrac{\tan x}{\tan {\pi/2}}}{1/\tan {\pi/2} -\tan x}$
or, $=\dfrac{\tan x \cot (\pi/2)+1}{\cot (\pi/2)-\tan x}$
or, $=\dfrac{-1}{\tan x}$
or, $=-\cot x$
