find tan p- cot R if PQ=12, PRcm =13cm , and angle Q =90
Answers
Step-by-step explanation:
given:
\text{In right angled $\triangle$PQR, PQ=12 cm and PR=13 cm}In right angled △PQR, PQ=12 cm and PR=13 cm
\textbf{To find:}To find:
tan\,P-cot\,RtanP−cotR
\textbf{Solution:}Solution:
\text{In right angled $\triangle$PQR, by Pythagoras theorem}In right angled △PQR, by Pythagoras theorem
\text{we get}we get
PR^2=PQ^2+QR^2PR
2
=PQ
2
+QR
2
13^2=12^2+QR^213
2
=12
2
+QR
2
169=QR^2+144169=QR
2
+144
\implies\,QR^2=25⟹QR
2
=25
\implies\,QR=5⟹QR=5
\text{Consider,}Consider,
tan\,P=\dfrac{QR}{PQ}tanP=
PQ
QR
\implies\bf\,tan\,P=\dfrac{5}{12}⟹tanP=
12
5
cot\,R=\dfrac{QR}{PQ}cotR=
PQ
QR
\implies\bf\,cot\,R=\dfrac{5}{12}⟹cotR=
12
5
\text{Now,}Now,
tan\,P-cot\,RtanP−cotR
=\dfrac{5}{12}-\dfrac{5}{12}=
12
5
−
12
5
=0=0
\therefore\textbf{The value of tan\,P-cot\,R is 0}∴The value of tanP-cotR is 0
Answer:
clearly, PQR is a right triangle with right angle at Q, hypotenuse as PR and one side as PQ
now
tanP - cotR
= QR/PQ - QR/PQ
= 0