Answer
Length using integral in polar coordinates:
$s = \tan A$
Length using trigonometry:
$\overline {PQ} = \tan A$
Work Step by Step
Write $r = \sec \theta = \frac{1}{{\cos \theta }}$. It follows that $r\cos \theta = 1$.
Since $x = r\cos \theta $. It implies that the polar equation $r = \sec \theta $ is a line $x=1$ in rectangular coordinates. From the figure attached, we see that the distance from the origin to a point in the line makes an angle $\theta$ with the $x$-axis.
We have
$r = f\left( \theta \right) = \sec \theta $, ${\ \ }$ $f'\left( \theta \right) = \sec \theta \tan \theta $
Using Eq.(7), we get the arc length for $0 \le \theta \le A$
$s = \mathop \smallint \limits_0^A \sqrt {{{\sec }^2}\theta + {{\sec }^2}\theta {{\tan }^2}\theta } {\rm{d}}\theta $
$s = \mathop \smallint \limits_0^A \sqrt {{{\sec }^2}\theta \left( {1 + {{\tan }^2}\theta } \right)} {\rm{d}}\theta $
Since ${\sec ^2}x - {\tan ^2}x = 1$, the integral becomes
$s = \mathop \smallint \limits_0^A \sqrt {{{\sec }^4}\theta } {\rm{d}}\theta = \mathop \smallint \limits_0^A {\sec ^2}\theta {\rm{d}}\theta $
From Eq. 14 of Section 8.2 (page 402) we have
$\smallint {\sec ^m}x{\rm{d}}x = \frac{{\tan x{{\sec }^{m - 2}}x}}{{m - 1}} + \frac{{m - 2}}{{m - 1}}\smallint {\sec ^{m - 2}}{\rm{d}}x$
So,
$s = \tan \theta |_0^A = \tan A$
Using trigonometry, we have the length segment from the points $P$ to $Q$ corresponding to $0 \le \theta \le A$ as
$\tan A = \frac{{\overline {PQ} }}{{\overline {OP} }}$
$\overline {PQ} = \tan A$
Thus, the two results agree.
