Answer
$\tan (0.2) \approx 0.2$
Work Step by Step
The local linear approximation of $f(x)$ at $x_0$ can be expressed as: $f(x) \approx f(x_0)+f^{\prime}(x_0)(x-x_0)$
Next, we have: $f(x)=\tan x$ at $x_0=0$
The local linear approximation of $f(x)$ at $x_0$ can be expressed as: $\tan (0.2) \approx \tan (0)+\sec^2 (0) (0.2 -0) =0.2$
Therefore, $\tan (0.2) \approx 0.2$
