Prove that, can anyone help me, please?

Question

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

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Answer 1 · 2017-12-11T02:14:12

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#