Question

In the given figure find tan P - cot R

Answer 1

Applying Pythagoras theorem for $ΔPQR,$ we obtain
$(PR)^ 2 = (PQ)^ 2 +( QR)^ 2$
$(13 \ cm) ^2 = (12\ cm)^ 2 + (QR)^ 2$
$169 \ cm^2 = 144 \ cm^2 + (QR)^ 2$
$25\ cm^2 = (QR)^ 2$
$QR = 5 \ cm$

\text{ tan P} = \frac{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠P }{\text{Side}\ \text{ Adjacent}\ \text{ to}\ ∠P}$$=\frac{QR}{PQ}=\frac{5}{12}
$\text{ cot R} = \frac{\text{Side}\ \text{Adjacent}\ \text{ to}\ ∠R }{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠R}$ $=\frac{QR}{PQ}=\frac{5}{12}$

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