Question

please answer my questions ​

Answer 1

Hii its tom85

using paythagoras theorem

k= H= sqrt (169 -25)= sqrt144=12

tan p - cot R

perpendicular/base (of angle p)- base/ perpendicular(of angle R)

5/ 12- 5/12

0

___________/\__________☺️

hope it may helps you dude

Answer 2

$\mathfrak{\underline{\underline{\green{Answer:-}}}}$

$\mathsf{tanP-cotR = 0}$

$\mathfrak{\underline{\underline{\green{Explanation:-}}}}$

Given:

Right angled Triangle [Refer the atachment]

In which,

PQ = 12cm

QR = 5cm

PR = 13cm

To Find:

Value of tanP - cotR

Solution:

Let us find the values of tanP and cotR

Value of tanP

As we know

$\boxed{\pink{ tan \theta = \dfrac{opposite \: side}{Adjacent \: side}}}$

Therefore

$\mathsf{ tanP = \dfrac{QR}{PQ}}$

$\mathsf{ tanP = \dfrac{5}{12}}$

Value of cotR

As we know

$\boxed{\pink{ cot \theta = \dfrac{Adjacent \: side}{Opposite \: side}}}$

Therefore

$\mathsf{ cotR = \dfrac{QR}{PQ}}$

$\mathsf{ cotR = \dfrac{5}{12}}$

Value of tanP -cotR

$\mathsf{ tanP-cotR = \dfrac{5}{12} - \dfrac{5}{12}}$

$\mathsf{ tanP-cotR = \cancel{\dfrac{5}{12}}- \cancel{\dfrac{5}{12}}}$

$\mathsf{ tanP-cotR = 0 }$

Hence,

$\mathsf{ tanP-cotR = 0 }$