Secant Function

Approach:

The secant is the reciprocal of the cosine, so it is undefined exactly at the zeros of the cosine, and its absolute value can never drop below $1$.

Steps:

1. By the unit-circle definition $\sec x = \frac{1}{\cos x}$, so the expression is undefined when $\cos x = 0$, that is when $x = \frac{\pi}{2} + k\pi$.
2. Since $|\cos x| \leq 1$ with equality exactly at $x = k\pi$, the value of $\frac{1}{\cos x}$ always satisfies $\left|\frac{1}{\cos x}\right| \geq 1$.
3. Both bounds are attained, so the range is $(-\infty, -1] \cup [1, +\infty)$.

Answer:

Domain: $x \neq \frac{\pi}{2} + k\pi$; Range: $(-\infty, -1] \cup [1, +\infty)$.

Approach:

Read the period off the addition formula for the cosine and check its parity by comparing $\sec(-x)$ with $\sec x$.

Steps:

1. The addition formula gives $\cos(x + 2\pi) = \cos x$, so $\sec(x + 2\pi) = \sec x$ and $2\pi$ is a period.
2. Computing $f(-x) = \sec(-x) = \frac{1}{\cos(-x)} = \frac{1}{\cos x} = f(x)$, the function is even.
3. Hence the period is $2\pi$ and the graph is symmetric about the $y$-axis.

Answer:

Period is $2\pi$, it is an even function.