Answer
a. $F(x)=\tan x-1$
b. $F'(x)=\sec^{2}x$
Work Step by Step
a.
$F(x)= \displaystyle \int_{\pi/4}^{x}\sec^{2}tdt=[\tan t]_{\pi/4}^{x}$
$F(x)=\displaystyle \tan x-\tan\frac{\pi}{4}$
$F(x)=\tan x-1$
b.
$\displaystyle \frac{d}{dx}[\tan x-1]=\sec^{2}x-0=\sec^{2}x$
