Question

If cotx=2mn/m^2-n^2 then cosx=?

Answer 1

cotx=B/P=2mn/m^2-n^2
cosx=B/H=2mn/H
cosx=2mn/√[(2mn)^2+(m^2-n^2)^2]
cosx=2mn/√(4m^2n^2+m^4+n^4-2m^2n^2)
cosx=2mn/√(m^4n^4-2m^2n^2)
cosx=2mn/m^2+n^2

Answer 2

cot x=adjacent side/opposite side
       =2mn/m²-n²
∴adjacent side=2mn
  opposite side=m²-n²
hypotenuse=√[(2mn)²+(m²-n²)²]
                  =√(4m²n²+m⁴+n⁴-2m²n²)
                  =√(m⁴+n⁴+2m²n²)
                  =√[(m²+n²)²]
                  =m²+n²
cos x=adjacent side/hypotenuse
        =2mn/m²+n²

Hope it helps u :-)