Question

plzz find:- tanP-cotR​

Answer 1

Answer:

p=12c.m.

h=13c.m

$b = \sqrt{ {13}^{2} - {12}^{2} } = 5c.m$

for tanP=QR/QP =5/12

cot R= 5/12

so,

tanP -tan Q =0

Answer 2

hope my answer helps you