In Fig. 8.13, find tan P – cot R.

Using Pythagoras theorem,

$P{R}^2 = Q{R}^2 + P{Q}^2$

Substituting the 12 for PQ and 13 for PR.

${13}^2 = Q{R}^2+{12}^2$

$169 = Q{R}^2+144$

$Q{R}^2 = 169-144$

$Q{R}^2 = 25$

$QR =\sqrt{25} = 5$

Thus, the side QR = 5 cm

As we know,

$\tan \left(P\right) =\frac{Opposite side}{Adjacent side}=\frac{QR}{PQ}=\frac{5}{12}$

$Cot \left(R\right) =\frac{Adjacent side}{Opposite side}=\frac{QR}{PQ}=\frac{5}{12}$

So,

$\tan \left(P\right) – cot \left(R\right)= \frac{5}{12}-\frac{5}{12} = 0$

Thus, $\tan \left(P\right) – cot \left(R\right)=0$