# In the figure, find tan P – cot R ?

## Solution :

In triangle PQR, we have

$$\angle$$ Q = 90,  PQ = 12 cm and PR = 13 cm

Using Pythagoras Theorem,

$${PR}^2$$ = $${PQ}^2$$ + $${QR}^2$$

$$\implies$$  $${QR}^2$$ = $${PR}^2$$ – $${PQ}^2$$

= $$(13)^2$$ – $$(12)^2$$ = 25

$$\implies$$  QR = 5 cm

We know that,

tan P = $$QR\over PQ$$ = tan P = $$5\over 12$$       ……..(1)

cot R = $$QR\over PQ$$  $$\implies$$  cot R = $$5\over 12$$            ………(2)

From (1) and (2), we have

tan P – cot R = $$5\over 12$$ – $$5\over 12$$ = 0