msinA =n cosA find tanA+cotA/tanA-cotA=
Answers
Answered by
0
msinA=ncosA
m/n.sinA/cosA=1
m/n.tanA=1
m/n=1/tanA
m/n=cotA
n/m=tanA
substituting in equation
(n/m+m/n)÷(n/m-m/n)
(n²+m²/mn)÷(n²-m²/mn)
n²+m²/mn×mn/n²-m²
(n²+m²)/(n²-m²)
[(n+m)(n+m)]/[(n+m)(n-m)]
=(n+m)/(n-m)
m/n.sinA/cosA=1
m/n.tanA=1
m/n=1/tanA
m/n=cotA
n/m=tanA
substituting in equation
(n/m+m/n)÷(n/m-m/n)
(n²+m²/mn)÷(n²-m²/mn)
n²+m²/mn×mn/n²-m²
(n²+m²)/(n²-m²)
[(n+m)(n+m)]/[(n+m)(n-m)]
=(n+m)/(n-m)
Similar questions