Question

if m cotA=n.then find the value of msinA-ncosA/mcosA+msinA?

Answer 1

(m sinA - n cosA)/(n cosA + m sinA)
 
Dividing the numerator (top) and denominator (bottom) by sinA
 
= [(m sinA - n cosA)/sinA]/[(n cosA + m sinA)/sinA]
= (m sinA/sinA  - n cosA/sinA)/(n cosA/sinA + m sinA/sinA)
= (m - ncotA)/(n cotA + m)
 
Since m cotA =n  or cotA = n/m, substituting it in the above expression, we get
 
= (m - n* n/m)/(n*n/m + m)
 
= [m - n^2/m]/[n^2/m + m]
 
= [(m^2 - n^2)/m]/[(n^2 + m^2)/m]
 
= [(m^2 - n^2)/m]  *[m/(n^2 + m^2)]
 
= (m^2 - n^2)/(n^2 + m^2)
 
Therefore, 
 
(m sinA - n cosA)/(n cosA + m sinA) =  (m^2 - n^2)/(m^2 + n^2)
 
I hope it helps!

Answer 2

Answer:

Step-by-step explanation:this is the easiest way to do these kind of questions if you understand what is done . Its all easy basics.