The function f(x)=sec[log(x+√(1+x^2))] is:

$f(x)=\sec(\log(x+\sqrt{1+x^2}))⇒f(-x)=\sec(\log(-x+\sqrt{1+x^2}))$

so $f(-x)=\sec(\log(-x+\sqrt{1+x^2}×\frac{x+\sqrt{1+x^2}}{x+\sqrt{1+x^2}})$

$f(-x)=\sec(\log(\frac{1}{\sqrt{1+x^2}+x})=\sec(-\log(x+\sqrt{1+x^2}))$

$f(-x)=\sec(\log(x+\sqrt{1+x^2}))$ even function