\lim\limits_{x\to 0} \dfrac{(1+x)^{\tan x}-1}{x^2}

=\lim\limits_{x\to 0} \dfrac{e^{\ln(1+x)^{\tan x}}-1}{x^2}

=\lim\limits_{x\to 0} \dfrac{\tan x\ln(1+x)}{x^2}

=\lim\limits_{x\to 0} \dfrac{\sin x}{x}\dfrac{\ln(1+x) }{x}

=1

So A is the right choice.

In the last step, we used two basic limits

\lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1

and

\lim\limits_{x\to 0} \dfrac{\ln(1+x) }{x} = 1

For the second one,

\lim\limits_{x\to 0} \dfrac{\ln(1+x) }{x}

= \lim\limits_{x\to 0} \ln(1+x)^{\frac{1}{x}}

= \lim\limits_{x\to 0} \ln e

=1