The function $\mathrm{f}(x)=\sec \left[\log \left(x+\sqrt{1+x^2}\right)\right]$ is _______ function

To determine the parity of $f(x)=\sec\left[\log\left(x+\sqrt{1+x^{2}}\right)\right]$, we evaluate $f(-x)$:

$f(-x) = \sec\left[\log\left(-x+\sqrt{1+(-x)^{2}}\right)\right] = \sec\left[\log\left(-x+\sqrt{1+x^{2}}\right)\right]$

Observe that the product $\left(x+\sqrt{1+x^{2}}\right)\left(-x+\sqrt{1+x^{2}}\right) = (1+x^{2}) - x^{2} = 1$, which implies:

$-x+\sqrt{1+x^{2}} = \frac{1}{x+\sqrt{1+x^{2}}}$

Substituting back:

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

Applying the logarithmic identity $\log(1/A) = -\log A$:

$f(-x) = \sec\left[-\log\left(x+\sqrt{1+x^{2}}\right)\right]$

Since secant is an even function $(\sec(-\theta) = \sec\theta)$, we obtain:

$f(-x) = \sec\left[\log\left(x+\sqrt{1+x^{2}}\right)\right] = f(x)$

This equality $f(-x) = f(x)$ confirms that the function is even.

$\boxed{A}$