Answer
$$\eqalign{
& f\left( x \right) = \tan x \cr
& {\text{The linear approximation at the point }}\left( {0,0} \right){\text{ is}} \to y = x \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \tan x,{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] \cr
& f'\left( x \right) = {\sec ^2}x \cr
& \cr
& {\text{The tangent line at the point }}\left( {c,f\left( c \right)} \right){\text{ is}} \cr
& y = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr
& {\text{We have the point }}\left( {0,0} \right) \cr
& y = 0 + {\sec ^2}\left( 0 \right)\left( {x - 0} \right) \cr
& y = 0 + 1\left( {x - 0} \right) \cr
& y = x \cr
& \cr
& {\text{Graph}} \cr} $$
