Question

if m cotA =n then the value of m SinA -ncosA/ncosA+msinA​

Answer 1

if doubt free to ask....

Answer 2

Answer:  (m²-n²)/(m²+n²)

Step-by-step explanation:

Given,

mcotA = n

mcosA/sinA = n

cosA/sinA=n/m

sinA/cosA = m/n

Multiplying with m/n in both sides,m

msinA/ncosA = m²/n²

Now,

Applying,  componendo - dividendo Rule,

(msinA - ncosA)/(msinA+ncosA) = (m²-n²)/(m²+n²)