Tangent, Cotangent, Secant & Cosecant Graphs

You’ll be able to

Locate the vertical asymptotes of each of the four non-sinusoidal trigonometric graphs
State the period of tangent and cotangent and explain why it is π rather than 2π
Describe the range of secant and cosecant and why no values lie between −1 and 1

Asymptotes live where the denominator vanishes

All four of these functions are quotients, so each has vertical asymptotes exactly where its denominator is zero. tan x = sin x/cos x and sec x = 1/cos x blow up where cos x = 0, at x = π/2 + nπ. cot x = cos x/sin x and csc x = 1/sin x blow up where sin x = 0, at x = nπ. Memorizing the asymptote locations is unnecessary if you remember which function sits in the denominator.

Asymptotes and periods

tan, sec: asymptotes at x = π/2 + nπ · cot, csc: asymptotes at x = nπ · period of tan and cot is π · period of sec and csc is 2π

n ranges over all integers. Tangent and cotangent are the only two with period π.

Why tangent repeats twice as fast

Advance x by π and both sine and cosine flip sign. Their quotient is therefore unchanged, because the two minus signs cancel: tan(x + π) = (−sin x)/(−cos x) = tan x. So tangent completes a full cycle in π. Secant and cosecant get no such cancellation — they are reciprocals of a single function, so the sign flip survives and their period stays 2π.

Worked example

Describe the graph of y = tan(x) on the interval (−π/2, π/2), then state what changes for y = 2·tan(x/2).

1.On (−π/2, π/2): tan x increases from −∞ to +∞, passing through the origin.
2.It has vertical asymptotes at both ends, x = ±π/2, and no maximum or minimum at all.
3.Its range is all real numbers, and it is odd: tan(−x) = −tan x.
4.For y = 2 tan(x/2): here b = 1/2, so the period becomes π/(1/2) = 2π.
5.Asymptotes stretch out to x = ±π, ±3π, …, i.e. odd multiples of π.
6.The factor 2 stretches vertically — but with an unbounded range that changes the steepness, not the extent.

Answer: y = tan x rises from −∞ to +∞ on each branch between consecutive asymptotes at odd multiples of π/2, with period π and range all reals. y = 2 tan(x/2) has period 2π with asymptotes at odd multiples of π, and is twice as steep at each crossing.

Period of a transformed tangent

y = a·tan(b(x − h)) + k has period π/|b|, not 2π/|b|

The single most common error in this topic. Tangent and cotangent use π in the numerator; the other four use 2π.

Checkpoint

What is the period of y = tan(3x)?

The forbidden band of secant and cosecant

Since |sin x| ≤ 1, its reciprocal satisfies |csc x| ≥ 1. So cosecant never takes a value strictly between −1 and 1: its range is (−∞, −1] ∪ [1, ∞). The same argument applies to secant. Graphically this produces the characteristic U shapes sitting above y = 1 and below y = −1, each U nestled into a peak or trough of the underlying sinusoid and reaching upward or downward toward the asymptotes on either side.

Tip

Sketch the sinusoid lightly first, then build its reciprocal on top. Zeros of the sinusoid become asymptotes; peaks at height 1 become minima of the U at height 1; troughs at −1 become maxima at −1. Every feature transfers by one rule.

Worked example

For y = sec(x), state the domain, range, period and asymptote locations.

1.sec x = 1/cos x, so it is undefined where cos x = 0.
2.cos x = 0 at x = π/2 + nπ, so those are the excluded values and the asymptotes.
3.Domain: all reals except x = π/2 + nπ.
4.Since |cos x| ≤ 1, |sec x| ≥ 1, so the range is (−∞, −1] ∪ [1, ∞).
5.cos has period 2π, and taking a reciprocal does not change how often values repeat.

Answer: Domain: all reals except odd multiples of π/2. Range: |y| ≥ 1. Period 2π. Asymptotes at x = π/2 + nπ. The value y = 1 is attained wherever cos x = 1, at x = 2nπ.

Checkpoint

Why does the graph of y = csc(x) have no points with y-values between −1 and 1?

Answer the 2 checkpoints as you read.