Answer
$L(x)=x$ and $Q(x)=x$
Work Step by Step
We have $f(x) = \tan x$
and $f'(x)= \sec^2 x ; \\f''(x)= 2 \sec^2 x \tan x$
$\implies f(0)=0; f'(0)=1; f''(0)=0$
Therefore, $L(x)=f(0)+xf'(0)=x$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=x$
So, $L(x)=x$ and $Q(x)=x$
