Trigonometric Identity

$\tan(A+\frac{\pi}{2})=\frac{\sin(A+\frac{\pi}{2})}{\cos(A+\frac{\pi}{2})}=\frac{\cos(A)}{-\sin(A)} =-\cot(A).$

$\tan (A+B)=\frac{\tan A + \tan B}{1-(\tan A \tan b)}$
In this case, $B=\frac {\pi}{2}$ ,
so,
$\tan [A+\frac {\pi}{2}]=\frac {\tan A + tan \frac{\pi}{2}}{1- \tan A \tan \frac{\pi}{2}}$
Dividing both numerator and denominator on RHS by ( $\tan \frac{\pi}{2}$ ),
$\Rightarrow \tan [A+\frac {\pi}{2}]= \frac {\frac{\tan A}{\tan \frac{\pi}{2}}+1}{\frac {1}{\tan \frac{\pi}{2}} - \tan A}$
The reason behind doing this is that the value of the trigonometric function $\tan$ at $\frac{\pi}{2}$ approaches $\infty$ , which would render the problem unsolvable. So, we divided both numerator and denominator by $\tan \frac{\pi}{2}$ , so that the problem can become solvable.
Now,
$\Rightarrow \tan [A+\frac {\pi}{2}]=\frac {\frac{\tan A}{\infty}+1}{\frac {1}{\infty} - \tan A}$
$\Rightarrow \tan [A+\frac {\pi}{2}]=\frac {0+1}{0 - \tan A}$
$\Rightarrow \tan [A+\frac {\pi}{2}]=\frac {1}{- \tan A}$
$\Rightarrow \tan [A+\frac {\pi}{2}]=-\cot A$
Hence, answer is $\boxed {-\cot A}$ .

As Tan(A +pi/2) lies in 2nd quadrant, tan changes to cot and as tan is -ve in 2nd quadrant, the answer is -cotA

$\tan (A + \frac{\pi}{2}) = \cot (\frac{\pi}{2} - (A + \frac{\pi}{2})) = \cot ( - A)=$ $= \frac{\cos ( - A)}{\sin ( -A)} = \frac{\cos (A)}{- \sin(A)} = - \cot (A)$