设 $f(x)$ 二阶可导, $f(1)=1, g(x)$ 为其反函数, $g^{\prime}(1)=g^{\prime \prime}(1)=a \neq 0$, 则 $\left.\left[\frac{\mathrm{d}^2}{\mathrm{~d} x^2} \int_0^{f(x)} \operatorname{tg}(t) \mathrm{d} t\right]\right|_{x=1}=$
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