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 \(=\frac{ 5}{12}-\frac{5}{12}=0\)