Answer
$${\text{ }}\tan x \approx x$$
Work Step by Step
$$\eqalign{
& \tan x \approx x,{\text{ }}{x_0} = 0 \cr
& {\text{Let }}f\left( x \right) = \tan x \cr
& {\text{ }}f\left( {{x_0}} \right) = f\left( 0 \right) = \tan 0 = 0 \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] \cr
& f'\left( x \right) = {\sec ^2}x \cr
& f'\left( {{x_0}} \right) = {\sec ^2}\left( 0 \right) = 1 \cr
& {\text{We can approximate values of }}f\left( x \right){\text{ by}} \cr
& {\text{ }}f\left( x \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \cr
& {\text{then}} \cr
& {\text{ }}\tan x \approx 0 + 1\left( {x - 0} \right) \cr
& {\text{ }}\tan x \approx x \cr} $$
