Answer
$0.0135$
Work Step by Step
The remainder in the n-th order Taylor's polynomial $T_n(x)$ for a function $f(x)$ centered at $a$ can be expressed as:
$|R_n(x)|=|f(x)-T_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}~~~(1)$
We are given the functions $f(x)=\tan x$
and $|f''(c)|=|4 \sec^2 c \tan^2 c+2 \sec^4 c | \leq 3 \implies M=3$
Put these values in the equation (1) to obtain:
$|R_{n}(x)| \leq 3\dfrac{|x-a|^{n+1}}{(n+1)!}$
Thus, we have: $|R_{2})| \leq 3 \dfrac{|x-0|^{2+1}}{(2+1)!}=0.0135$