What Does Sec 0 Degrees Mean?
Secant is the reciprocal of cosine, $\sec\theta = \dfrac{1}{\cos\theta}$, and it is one of the six secant-family functions. In a right triangle it is the hypotenuse divided by the adjacent side.
On the unit circle, a circle of radius $1$ centred at the origin, secant is $\frac{1}{x}$, the reciprocal of the $x$-coordinate. At $0^\circ$ that point is $(1, 0)$, so $\sec 0^\circ = \frac{1}{1} = 1$. This mirrors cos 0 degrees, which is also $1$, since a number and its reciprocal are equal only when that number is $1$.
Where Does Sec 0 Degrees Show Up?
The value marks a line pointing straight along the horizontal, where the angle of elevation is zero. Secant compares the hypotenuse to the adjacent side, and at $0^\circ$ those two lengths coincide, giving a ratio of $1$.
It appears as the starting value of the secant curve, the point from which the graph rises to its asymptote at $90^\circ$. In surveying and optics, a sightline held perfectly level carries a secant of $1$, the baseline against which every steeper angle is measured.
What Is The Value Of Sec 0 Degrees In The Secant Table?
Secant is at its smallest, exactly $1$, when the angle is $0^\circ$, then climbs without bound toward $90^\circ$.
Angle (degrees) | Angle (radians) | $\sec\theta$ (exact) | $\sec\theta$ (decimal) |
|---|---|---|---|
$0^\circ$ | $0$ | $1$ | $1.0000$ |
$30^\circ$ | $\dfrac{\pi}{6}$ | $\dfrac{2}{\sqrt{3}}$ | $1.1547$ |
$45^\circ$ | $\dfrac{\pi}{4}$ | $\sqrt{2}$ | $1.4142$ |
$60^\circ$ | $\dfrac{\pi}{3}$ | $2$ | $2.0000$ |
$90^\circ$ | $\dfrac{\pi}{2}$ | undefined | — |
Secant never sits between $-1$ and $1$, so $1$ is the lowest value it reaches in the first quadrant. That floor happens at $0^\circ$, where cosine is at its peak of $1$ and the reciprocal flips it back to $1$.
How Do You Find The Exact Value Of Sec 0 Degrees?
Two routes both give $1$, and neither needs a triangle drawn at $0^\circ$, which would collapse flat.
Method 1: The reciprocal of cosine.
$$\sec 0^\circ = \frac{1}{\cos 0^\circ}$$
$$\cos 0^\circ = 1$$
$$\sec 0^\circ = \frac{1}{1} = 1$$
Method 2: The unit circle.
At $0^\circ$ the radius lies along the positive $x$-axis and meets the circle at $(1, 0)$. Secant reads the reciprocal of the $x$-coordinate:
$$\sec 0^\circ = \frac{1}{x\text{-coordinate}} = \frac{1}{1} = 1$$
Both agree because the $x$-coordinate on the unit circle is exactly $\cos\theta$, so $\frac{1}{x}$ and $\frac{1}{\cos\theta}$ are the same reading.
Examples Of Sec 0 Degrees
Example 1
Evaluate $7\sec 0^\circ$.
$$7\sec 0^\circ = 7 \times 1 = 7$$
Example 2
Find $\sec 0^\circ$ from its reciprocal definition.
Wrong attempt. A student writes $\sec 0^\circ = \text{undefined}$, reasoning that reciprocal functions blow up at $0^\circ$ the way cosecant does.
That breaks: cosecant is $\frac{1}{\sin\theta}$, and $\sin 0^\circ = 0$, so csc $0^\circ$ really is undefined. Secant is $\frac{1}{\cos\theta}$, and $\cos 0^\circ = 1$, which is nowhere near zero.
Correct. $\sec 0^\circ = \dfrac{1}{\cos 0^\circ} = \dfrac{1}{1} = 1$. The function that is undefined at $0^\circ$ is cosecant, not secant.
Example 3
Evaluate $\sec 0^\circ + \cos 0^\circ$.
$$\sec 0^\circ + \cos 0^\circ = 1 + 1 = 2$$
Example 4
Simplify $5\sec 0^\circ - 3\tan 0^\circ$.
$$5(1) - 3(0) = 5 - 0 = 5$$
Example 5
Verify the identity $\sec^2 0^\circ - \tan^2 0^\circ = 1$.
$$\sec^2 0^\circ - \tan^2 0^\circ = (1)^2 - (0)^2 = 1 - 0 = 1$$
The Pythagorean identity holds, as it must for every angle where secant is defined.
Where Students Trip Up On Sec 0 Degrees
Mistake 1: Calling sec 0° undefined
Where it slips in: Grouping secant with cosecant, which is undefined at $0^\circ$.
Don't do this: Writing $\sec 0^\circ = \text{undefined}$.
The correct way: Secant is $\frac{1}{\cos\theta}$, and $\cos 0^\circ = 1$, so $\sec 0^\circ = 1$. The first instinct is to expect a huge or undefined number, because secant does blow up at $90^\circ$; at $0^\circ$ it is at its smallest, exactly $1$.
Mistake 2: Reading sec 0° as 0
Where it slips in: Sliding from "sec" to "sine" and using $\sin 0^\circ = 0$.
Don't do this: Writing $\sec 0^\circ = 0$.
The correct way: Secant reciprocates cosine, not sine. Since $\cos 0^\circ = 1$, its reciprocal is $1$, and $0$ never appears as a secant value at all.
Mistake 3: Flipping the reciprocal onto sine
Where it slips in: Recall that pairs secant with the wrong base function.
Don't do this: Writing $\sec 0^\circ = \dfrac{1}{\sin 0^\circ} = \dfrac{1}{0}$.
The correct way: Secant is $\frac{1}{\cos\theta}$; the $\frac{1}{\sin\theta}$ form is cosecant. Keeping the pairing straight, secant with cosine, keeps $\sec 0^\circ$ at a clean $1$.
Key Takeaways
Sec 0 degrees equals $1$, because $\sec\theta = \dfrac{1}{\cos\theta}$ and $\cos 0^\circ = 1$.
On the unit circle the point at $0^\circ$ is $(1, 0)$, and $\frac{1}{x} = \frac{1}{1} = 1$.
Secant is defined at $0^\circ$; it is cosecant that is undefined there, so do not confuse the two.
In degrees or radians the value is the same: $\sec 0^\circ = \sec 0 = 1$, the smallest value secant ever reaches.
To take sec 0 degrees and the reciprocal functions further with a teacher, explore Bhanzu's trigonometry tutor, high school math tutor, or online math classes.
Practice These To Solidify Your Understanding
Evaluate $4\sec 0^\circ - 2\cos 0^\circ$.
Show that $\sec 0^\circ \times \cos 0^\circ = 1$.
Explain why csc $0^\circ$ is undefined while $\sec 0^\circ = 1$.
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