Answer
$tan~x~\rightarrow+\infty$
$cot~x~\rightarrow0^+$
Work Step by Step
As $x\rightarrow(\frac{\pi}{2})^-$, $sin~x\rightarrow1^-$ and $cos~x\rightarrow0^+$
Hence, as $x\rightarrow(\frac{\pi}{2})^-$:
$tan~x\rightarrow\frac{1^-}{0^+}=+\infty$
$cot~x\rightarrow\frac{0^+}{1^-}=0^+$
