In the given figure, find tan P – cot R.

From figure,

PQR is a right angled triangle.

By pythagoras theorem, we get :

⇒ PR² = PQ² + QR²

⇒ 13² = 12² + QR²

⇒ QR² = 169 - 144

⇒ QR² = 25

⇒ QR = $\sqrt{25}$ = 5.

tan P = $\dfrac{\text{Side opposite to ∠P}}{\text{Side adjacent to ∠P}} = \dfrac{QR}{PQ} = \dfrac{5}{12}$.

cot R = $\dfrac{\text{Side adjacent to ∠R}}{\text{Side opposite to ∠R}} = \dfrac{QR}{PQ} = \dfrac{5}{12}$.

Substituting values in tan P - cot R, we get :

$\Rightarrow \dfrac{5}{12} - \dfrac{5}{12}$ = 0.

Hence, tan P - cot R = 0.