In Fig. find tan P – cot R.

By using the Pythagoras theorem, we have

\( \Rightarrow PR^2 \ = \ PQ^2 \ + \ QR^2 \)
\(\Rightarrow 13^2 \ = \ 12^2 \ + \ QR^2 \)
\( \Rightarrow QR^2 \ = \ 13^2 \ - \ 12^2 \ \)
\( \Rightarrow QR^2 \ = \ 169 \ - \ 144 \ = \ 25 \)
\( \Rightarrow QR \ = \ \sqrt{25} \ = \ 5 \)

\(\therefore \) \( tanP \ = \ \frac{P}{B} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12} \)

\( cotR \ = \ \frac{B}{P} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12} \)

Hence, \( tan P \ – \ cot R \ = \ \frac{5}{12} \ - \ \frac{5}{12} \ = \ 0 \)