Prove that, can anyone help me, please?

cosh[ln(x+sqrt((x^2-1)]] = x

1 Answer
Dec 11, 2017

By definition:

coshx =(e^x+e^-x)/2

so:

coshln(x+sqrt(x^2-1)) = 1/2(e^(ln(x+sqrt(x^2-1)))+e^(-ln(x+sqrt(x^2-1))))

coshln(x+sqrt(x^2-1)) = 1/2(x+sqrt(x^2-1)+1/(x+sqrt(x^2-1)))

coshln(x+sqrt(x^2-1)) = 1/2((x+sqrt(x^2-1))^2+1)/(x+sqrt(x^2-1))

coshln(x+sqrt(x^2-1)) = 1/2((x^2+2xsqrt(x^2-1)+x^2-1+1)/(x+sqrt(x^2-1)))

coshln(x+sqrt(x^2-1)) = 1/2((2x^2+2xsqrt(x^2-1))/(x+sqrt(x^2-1)))

coshln(x+sqrt(x^2-1)) =x(x+sqrt(x^2-1))/(x+sqrt(x^2-1))

coshln(x+sqrt(x^2-1)) =x