Answer
$${\text{True}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\arctan \left( {\tan x} \right)} \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {\arctan u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}} \cr
& \frac{d}{{dx}}\left[ {\arctan \left( {\tan x} \right)} \right] = \frac{1}{{1 + {{\left( {\tan x} \right)}^2}}}\frac{d}{{dx}}\left[ {\tan x} \right] \cr
& {\text{Differentiating}} \cr
& \frac{d}{{dx}}\left[ {\arctan \left( {\tan x} \right)} \right] = \frac{1}{{1 + {{\tan }^2}x}}\left( {{{\sec }^2}x} \right) \cr
& {\text{Use the pythagorean identity }}{\sec ^2}x = {\tan ^2}x + 1 \cr
& \frac{d}{{dx}}\left[ {\arctan \left( {\tan x} \right)} \right] = \frac{1}{{1 + {{\tan }^2}x}}\left( {{{\tan }^2}x + 1} \right) \cr
& \frac{d}{{dx}}\left[ {\arctan \left( {\tan x} \right)} \right] = 1 \cr
& {\text{The statement is true}}{\text{.}} \cr} $$
