Applying Pythagoras theorem for $\triangle PQR$, we obtain

$PR ^{2}= PQ ^{2}+ QR ^{2}$

$(13\, cm )^{2}=(12\, cm )^{2}+ QR ^{2}$

$169 \,cm ^{2}=144 \,cm ^{2}+ QR ^{2}$

$25\, cm ^{2}= QR ^{2}$

$QR =5\, cm$

$\tan P=\frac{\text { Side opposite to } \angle P }{\text { Side adjacent to } \angle P }=\frac{ QR }{ PQ }$

$=\frac{5}{12}$

$\cot R=\frac{\text { Side adjacent to } \angle R }{\text { Side opposite to } \angle R }=\frac{ QR }{ PQ }$

$=\frac{5}{12}$

$\tan P- \cot R =\frac{5}{12}-\frac{5}{12}=0$