Tangent and the meaning of a vertical asymptote

Explain it like I am five

Picture cutting 1 pizza into slices: slices of 0.1 pizza make 1 ÷ 0.1 = 10 slices, slices of 0.01 make 100. Tiny divisors give huge answers, and dividing by exactly 0 gives no answer.

Tangent is tan x = $\frac{\sin x}{\cos x}$ . As x creeps up toward $π over 2$ ≈ 1.5708 (≈ means about), sin x nears 1 while cos x shrinks toward 0. A calculator in radians shows it: tan 1.5 ≈ 14.10, tan 1.55 ≈ 48.08, tan 1.57 ≈ 1255.77. The heights pass any ceiling you name, so they are unbounded. Right after $π over 2$ , cos x is tiny and negative, so the heights are hugely negative. At x = $π over 2$ itself cos x = 0, and there is no height at all.

So the dashed line x = $π over 2$ is a vertical asymptote: a vertical line the graph hugs while its heights grow without bound, with no point on it. Tangent has one wherever cos x = 0, at x = $π over 2$ + nπ, and each piece between two of them, a branch, rises from left to right.

In plain words

Imagine walking along a ramp that climbs beside a tall fence. You can move closer to the fence while your height becomes greater than any height you name. The fence marks an input to watch, not a height you reach. A vertical asymptote works that way: nearby outputs grow without bound on at least one side of a vertical line. Unbounded means the outputs pass every fixed ceiling or every fixed floor. Tangent has separate pieces called branches. Each branch is continuous, with no break inside it, and rises as you move right. Its excluded boundary inputs separate those pieces.

These original inputs in radians approach ≈ 1.570796 from the left; the highlighted column still has a finite output.

Reminder

Coordinates and plotting. An ordered pair is (input, output): tan( $π over 4$ ) = 1 gives ( $π over 4$ , 1), with the horizontal coordinate first.
Radian fractions and common denominators. To copy a point by π, use π = $4 π over 4$ : − $π over 4$ + π = $3 π over 4$ .
Integer multiples. An integer can be negative, zero, or positive: n = −1 in $π over 2$ + nπ gives − $π over 2$ .
Unit-circle values. The circle point lists (cos x, sin x): at $π over 2$ it is (0, 1), so tangent divides 1 by 0.
Fraction division and signs. Equal nonzero numbers divide to 1; a negative divided by a positive is negative: − $square root of 2 over 2$ ÷ $square root of 2 over 2$ = −1.
Zero numerator and zero denominator. $0 over 1$ = 0 is allowed; $1 over 0$ is undefined because no real number multiplied by 0 gives 1.
Inequalities and interval endpoints. (− $π over 2$ , $π over 2$ ) means − $π over 2$ < x < $π over 2$ , with both endpoints excluded.
Substitution checks. After finding a forbidden input, plug it back into the denominator: cos( $π over 2$ ) = 0 verifies the exclusion.
Quadrant signs. Quadrant I is upper right, with positive sine and cosine. Quadrant II is upper left, with positive sine and negative cosine.

Why it works. On the unit circle, tan x is the vertical coordinate divided by the horizontal coordinate. Near from the left, sine approaches 1 while positive cosine shrinks toward 0. Dividing a number near 1 by a very small positive number makes a very large positive output. From the right, cosine is small and negative, so the output is very negative. At itself the denominator is zero, so there is no output. Infinity describes unbounded behavior; it is not a value of tangent.

RuleRule: tan x = , with x ≠ + nπ for every integer n. The vertical asymptotes are x = + nπ.
Each branch is continuous and increasing; the central branch is on (−, ) and the pattern repeats every π.

The same idea, five ways

Say it

Say: a vertical asymptote is a vertical line near which the outputs become unbounded on at least one side.

Write it

Write: x = $π over 2$ is a vertical asymptote of tangent because its nearby outputs become arbitrarily large in magnitude.

In math

x = $π over 2$
x < $π over 2$ , x approaches $π over 2$ : tan x grows without bound
x > $π over 2$ , x approaches $π over 2$ : tan x becomes negative without bound
tan( $π over 2$ ) is undefined
∞ is not a real output

You can approach a fence while climbing beyond every height you name; the fence is an input boundary, not the top of the ramp.

See it

The x-coordinates are in radians; the branch rises between the dashed lines, and nearby pieces lie on the other sides.

The same idea, other ways

As a graph

A vertical asymptote is a vertical line, not a point with an output. The rising central branch runs between x = $−π over 2$ and x = $π over 2$ . It leaves any fixed drawing window near either end, so the edge of the picture is not an endpoint of the function.

In radians, the central branch leaves the visible window as it approaches its two boundary lines.

As smaller portions

One whole divided into tenths gives 10 portions. Divided into hundredths it gives 100 portions. Tangent near $π over 2$ behaves similarly because its numerator stays near 1 while its positive denominator becomes tiny. No calculation ever divides by zero or produces an actual infinity.

1 ÷ 0.1 = 10

1 ÷ 0.01 = 100

1 ÷ 0.001 = 1000

Division by 0 is undefined

Smaller positive divisors give larger finite quotients; zero is still excluded.

With nearby inputs

Read the column under 1.4: its output is about 5.7979. The column under 1.57 has output about 1255.7656. Both inputs are less than $π over 2$ ≈ 1.570796. They show the approach from the left. Each output is rounded to four decimal places; the table illustrates growth near the boundary, while the shrinking-denominator argument explains why it continues without bound.

Compare the highlighted columns in this original table of radian inputs, with outputs rounded to four decimal places.

As a turning ray

In the central interval, the unit-circle point is on the right half of the circle. The ray from the origin turns from almost straight down to horizontal to almost straight up. Its vertical coordinate divided by its positive horizontal coordinate is tangent. That ratio increases throughout the interval, matching the rising graph.

At radians, equal positive vertical and horizontal coordinates give tangent 1.

x in radians	tan x, rounded to four decimal places
1.4	5.7979
1.5	14.1014
1.55	48.0785
1.56	92.6205
1.57	1255.7656

.1Vertical asymptote: nearby outputs become unbounded

A one-sided approach means moving toward an input while staying entirely on its left or entirely on its right. You can approach a doorway from either side without standing in the doorway. Near a vertical asymptote, watch the outputs along that approach. For tangent at $π over 2$ , the left-side outputs grow positively without bound and the right-side outputs grow negatively without bound. That statement concerns nearby accepted inputs. It does not give tangent an output at $π over 2$ .

One-sided approach: stay on one specified side of the input.
At $π over 2$ : tangent grows positively without bound from the left and negatively without bound from the right.
Infinity: a description of unbounded growth, not a real output.
Not continuous (discontinuous): the complete tangent picture has breaks at its excluded inputs.
Continuous: within each branch, the tangent curve has no break.

The vertical line is at radians; neither neighboring piece gives an output on that line.

The same idea, five ways

Say it

Say: approach $π over 2$ from the left means use smaller and smaller gaps below $π over 2$ ; from the right means use gaps above it.

Write it

Write: tangent has opposite unbounded behavior on the two sides of $π over 2$ , and it has no value at $π over 2$ .

In math

x < $π over 2$
x > $π over 2$
x ≠ $π over 2$
tan( $π over 2$ ) is undefined
Graph words: unbounded above from the left, unbounded below from the right

Approach a doorway along the hall from its left side or its right side; the direction of approach matters.

See it

Around radians, the left piece rises out of the window while the right piece falls out of it.

Worked exampleA zero denominator and nearby finite values

Explain tangent at $π over 2$ , at 1.56, and at 1.58 radians. This asks you to separate the excluded input from accepted inputs on either side of it.

The middle column is excluded; the nearby radian inputs have finite outputs with opposite signs.

At x = $π over 2$ , write tan x = $1 over 0$ .The unit-circle values are sin( $π over 2$ ) = 1 and cos( $π over 2$ ) = 0; this substitution checks the excluded input.
Reject $1 over 0$ as undefined.No real output multiplied by 0 gives 1.
Since 1.56 < 1.570796 < 1.58, classify 1.56 as left-side and 1.58 as right-side. $π over 2$ is approximately 1.570796 radians, so the inequalities identify the direction of each approach.
Read the two columns: tan(1.56) ≈ 92.6205 and tan(1.58) ≈ −108.6492.Both denominators are nonzero, so both accepted inputs have finite outputs; these outputs are rounded to four decimal places.

Answer

tan( $π over 2$ ) is undefined.
tan(1.56) ≈ 92.6205, on the left.
tan(1.58) ≈ −108.6492, on the right.

Check The supplied coordinate approximations provide a numerical cross-check: at 1.56, sine ≈ 0.9999417202 and cosine ≈ 0.0107961171 give ≈ 92.6205. At 1.58, sine ≈ 0.9999576465 and cosine ≈ −0.0092035433 give ≈ −108.6492. These coordinates are rounded to ten decimal places and supplied for illustration. Your independent circle check is the sign: sine is positive on both sides, cosine is positive on the left and negative on the right. At , the exact cosine is 0, so that input is still excluded.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: tan() = ∞.

Infinity is not a real output, and the quotient at this exact input divides by zero.

✓ Instead: tan() is undefined; nearby left-side outputs grow without bound.

Tips and tricks

Tip: write the side of the approach before deciding whether outputs grow positively or negatively.

.2One tangent branch: three points and two boundary lines

A branch is one connected piece of the graph. Think of one stretch of a road between two closed gates. You may travel inside that stretch without jumping a gap. That is what continuous means here. Increasing means that moving to a larger input within this branch gives a larger output. The central tangent branch starts below every fixed height near its left gate, passes through the origin, and climbs above every fixed height near its right gate. The gates are the excluded inputs $−π over 2$ and $π over 2$ .

Branch: one continuous piece between consecutive vertical asymptotes.
Increasing: larger inputs within the same branch give larger outputs.
Central interval: (− $π over 2$ , $π over 2$ ), with both endpoints excluded.
The central tangent branch passes through the origin, so (0, 0) is both an x-intercept and a y-intercept.

On the central radian interval, the graph rises smoothly through the three points.

The same idea, five ways

Say it

Say: one branch is one unbroken piece; increasing means farther right gives higher outputs on that piece.

Write it

Write: the central tangent branch rises continuously between its two excluded boundary inputs.

In math

− $π over 2$ < x < $π over 2$
(− $π over 2$ , $π over 2$ )
{x | − $π over 2$ < x < $π over 2$ }
If − $π over 2$ < $x sub 1$ < $x sub 2$ < $π over 2$ , then tan $x sub 1$ < tan $x sub 2$ .
Graph words: one rising branch through the origin

Follow one climbing road between two closed gates without jumping any gap inside the road.

See it

The x-coordinates are radians; the three marked points guide a single continuous rising branch.

Worked exampleThe tangent branch through (−π, 0)

Graph the branch of y = tan x that passes through the point (−π, 0). The three-column table gives exact values on this branch: tan(− $4 π over 3$ ) = − $square root of 3$ , tan(−π) = 0 and tan(− $2 π over 3$ ) = $square root of 3$ . The table lists each input as a count of thirds of π, so −4, −3 and −2 stand for − $4 π over 3$ , −π and − $2 π over 3$ . It shows $square root of 3$ rounded to 1.732. Find and check the two boundary lines of this branch, plot the three points and draw the branch. Then copy the branch π units right and π units left.

Three points on the branch of y = tan x through (−π, 0). Inputs are counted in thirds of π, so −4, −3 and −2 mean −, −π and −. The outputs −, 0 and are shown rounded as −1.732, 0 and 1.732.

Solve cos x = 0: x = $π over 2$ + nπ. The solutions just on either side of −π come from n = −2, giving x = $π over 2$ − 2π = − $3 π over 2$ , and n = −1, giving x = $π over 2$ − π = − $π over 2$ . Since − $3 π over 2$ < −π < − $π over 2$ and the two are exactly π apart, no other excluded input lies between them. So this branch lives on − $3 π over 2$ < x < − $π over 2$ .tan x = $\frac{\sin x}{\cos x}$ , so every input with cos x = 0 makes the denominator zero and must be excluded. Two neighboring excluded inputs are the boundary lines of one branch.
Check both boundary inputs on the unit circle. − $3 π over 2$ is coterminal with − $3 π over 2$ + 2π = $π over 2$ , the point (0, 1), so cos(− $3 π over 2$ ) = 0 and sin(− $3 π over 2$ ) = 1. − $π over 2$ is the point (0, −1), so cos(− $π over 2$ ) = 0 and sin(− $π over 2$ ) = −1. Now draw the dashed vertical lines x = − $3 π over 2$ and x = − $π over 2$ .A boundary line is drawn only after the denominator is confirmed to be zero there. At each line the numerator is 1 or −1, not 0. So next to each line, tan x is a nonzero number divided by a number close to zero. There is no value on the line itself, and the values beside it grow without bound.
Confirm the table values as sin x ÷ cos x. − $4 π over 3$ is coterminal with $2 π over 3$ , the point (− $1 over 2$ , $square root of 3 over 2$ ), so tan(− $4 π over 3$ ) = $square root of 3 over 2$ ÷ (− $1 over 2$ ) = − $square root of 3$ . −π is the point (−1, 0), so tan(−π) = 0 ÷ (−1) = 0. − $2 π over 3$ is coterminal with $4 π over 3$ , the point (− $1 over 2$ , − $square root of 3 over 2$ ), so tan(− $2 π over 3$ ) = (− $square root of 3 over 2$ ) ÷ (− $1 over 2$ ) = $square root of 3$ . Plot (− $4 π over 3$ , − $square root of 3$ ), (−π, 0) and (− $2 π over 3$ , $square root of 3$ ), placing $square root of 3$ at about 1.732.Exact unit-circle values give the correct heights. The table's −4, −3 and −2 are thirds of π, so they are exactly these three inputs. All three inputs lie strictly between the boundary lines, so they belong to this branch: − $3 π over 2$ = − $9 π over 6$ < − $8 π over 6$ = − $4 π over 3$ , and − $2 π over 3$ = − $4 π over 6$ < − $3 π over 6$ = − $π over 2$ .
Draw one smooth rising curve through the three points, kept between the dashed lines. As x approaches − $3 π over 2$ from the right, the curve drops without bound. As x approaches − $π over 2$ from the left, it rises without bound. It never touches either line.Each branch is continuous and increasing. On − $3 π over 2$ < x < − $π over 2$ , the matching unit-circle points lie on the left half of the circle, so cos x is negative there. Near x = − $3 π over 2$ , tan x is about 1 divided by a tiny negative number, which is a large negative value. Near x = − $π over 2$ , it is about −1 divided by a tiny negative number, which is a large positive value.
Add π to every input for the right copy and subtract π for the left copy. Right copy: dashed lines x = − $π over 2$ and x = $π over 2$ , points (− $π over 3$ , − $square root of 3$ ), (0, 0) and ( $π over 3$ , $square root of 3$ ). Left copy: dashed lines x = − $5 π over 2$ and x = − $3 π over 2$ , points (− $7 π over 3$ , − $square root of 3$ ), (−2π, 0) and (− $5 π over 3$ , $square root of 3$ ). Draw each copy as its own rising branch between its two lines.The tangent pattern repeats every π. A point moved π units left or right is still on the graph at the same height, and each boundary line x = $π over 2$ + nπ moves onto the next one. Neighboring branches share the lines x = − $3 π over 2$ and x = − $π over 2$ . Those inputs are excluded, so the branches stay separate and no curve crosses a dashed line.

Answer

The boundary lines (vertical asymptotes) are x = − and x = −. The branch is the continuous increasing curve on − < x < − through (−, −), (−π, 0) and (−, ). It falls without bound as x approaches − from the right and rises without bound as x approaches − from the left. Copied π units right, it is the branch on − < x < through (−, −), (0, 0) and (, ). Copied π units left, it is the branch on − < x < − through (−, −), (−2π, 0) and (−, ).

Check Recompute two copied points directly. tan() = ÷ = , the same height as at −. tan(−) = (−) ÷ = −, the same height as at −. The zero sits halfway between the boundary lines: (− − ) ÷ 2 = −2π ÷ 2 = −π. In decimals, − ≈ −4.712 < − ≈ −4.189 < −π ≈ −3.142 < − ≈ −2.094 < − ≈ −1.571. The heights −1.732, 0 and 1.732 increase in the same order, so all three points are inside the branch and the curve rises. Finally, cos(−) = cos(−) = − ≠ 0, so no plotted input is an excluded one.

Work to write

cos x = 0 when x = $π over 2$ + nπ; on either side of −π this gives x = − $3 π over 2$ (n = −2) and x = − $π over 2$ (n = −1).
cos(− $3 π over 2$ ) = 0 and cos(− $π over 2$ ) = 0, so x = − $3 π over 2$ and x = − $π over 2$ are the dashed boundary lines (vertical asymptotes).
tan(− $4 π over 3$ ) = $square root of 3 over 2$ ÷ (− $1 over 2$ ) = − $square root of 3$ , tan(−π) = 0 ÷ (−1) = 0, tan(− $2 π over 3$ ) = (− $square root of 3 over 2$ ) ÷ (− $1 over 2$ ) = $square root of 3$ .
Points plotted: (− $4 π over 3$ , − $square root of 3$ ), (−π, 0), (− $2 π over 3$ , $square root of 3$ ).
One increasing branch on − $3 π over 2$ < x < − $π over 2$ : it falls without bound toward x = − $3 π over 2$ and rises without bound toward x = − $π over 2$ , never touching either line.
Copies: π right on − $π over 2$ < x < $π over 2$ through (0, 0); π left on − $5 π over 2$ < x < − $3 π over 2$ through (−2π, 0); each copy has its own boundary lines.

The boundary lines (vertical asymptotes) are x = − $3 π over 2$ and x = − $π over 2$ . The branch is the continuous increasing curve on − $3 π over 2$ < x < − $π over 2$ through (− $4 π over 3$ , − $square root of 3$ ), (−π, 0) and (− $2 π over 3$ , $square root of 3$ ). It falls without bound as x approaches − $3 π over 2$ from the right and rises without bound as x approaches − $π over 2$ from the left. Copied π units right, it is the branch on − $π over 2$ < x < $π over 2$ through (− $π over 3$ , − $square root of 3$ ), (0, 0) and ( $π over 3$ , $square root of 3$ ). Copied π units left, it is the branch on − $5 π over 2$ < x < − $3 π over 2$ through (− $7 π over 3$ , − $square root of 3$ ), (−2π, 0) and (− $5 π over 3$ , $square root of 3$ ).

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: connect the central branch to the next branch across x = .

At there is no tangent output, and the neighboring branches have opposite unbounded behavior near that line.

✓ Instead: Keep the branches separate; draw a dashed vertical asymptote at x = .

Tips and tricks

Tip: place the zero between the two boundary lines before drawing the curve.

.3Complete tangent graph: repeat by π

Think of wallpaper made from one repeated panel. Once you draw one panel correctly, you can place identical panels beside it. Tangent works the same way: its period is π, so an input moved π units left or right keeps the same output. The copying refers to pieces of the parent graph, not to making a new function. Every copied branch still rises. Between neighboring branches, the excluded input gives a vertical asymptote. The complete graph is not continuous across those excluded inputs, even though every individual branch is continuous.

Period: π.
Zeros: x = nπ, because sine is 0 and cosine is 1 or −1 there.
Vertical asymptotes: x = $π over 2$ + nπ, because cosine is 0 there.
Here n is an integer, including negative integers and 0.

The radian graph repeats rising branches and excludes every dashed vertical line.

The same idea, five ways

Say it

Say: move one branch π units sideways and its heights repeat.

Write it

Write: the complete tangent graph repeats each continuous rising branch every π radians.

In math

tan(x + π) = tan x
tan(x + nπ) = tan x, n an integer
x = $π over 2$ + nπ: vertical asymptotes
x = nπ: zeros
Graph words: repeated rising branches

Identical wallpaper panels repeat at a fixed spacing, with a seam between panels.

See it

With the horizontal axis in radians, consecutive rising branches repeat π units apart.

Worked exampleThe next tangent branch to the right

Sketch the tangent branch on ( $π over 2$ , $3 π over 2$ ). This asks you to copy the central branch one period to the right while keeping all its heights.

These copied radian inputs keep the central branch's three output heights.

The branch is a copy π radians to the right of the central branch.

Move the left boundary from − $π over 2$ to − $π over 2$ + π = $π over 2$ , and the right boundary to $π over 2$ + π = $3 π over 2$ .A period of π moves every input π units right without changing its output behavior.
Add π = $4 π over 4$ to − $π over 4$ , 0, and $π over 4$ , giving $3 π over 4$ , π, and $5 π over 4$ .Fractions with the same denominator add by their numerators; the copied inputs remain in the same left-to-right order.
Keep the outputs −1, 0, and 1, then draw a rising branch through the corresponding table columns.Tangent's period preserves each output, and the copied branch has the same shape.
Check both boundaries in cosine: cos( $π over 2$ ) = 0 and cos( $3 π over 2$ ) = 0.Substitution verifies that the copied boundary inputs still make the original denominator zero and must be excluded.

Answer

Interval: ( $π over 2$ , $3 π over 2$ ).
Boundary lines: x = $π over 2$ , x = $3 π over 2$ .
Points: ( $3 π over 4$ , −1), (π, 0), ( $5 π over 4$ , 1).
Shape: one continuous rising branch.

Check Check the points directly on the unit circle: at sine and cosine have equal magnitudes and opposite signs, giving −1; at π the numerator is 0 and the denominator is −1; at both coordinates are equal and negative, giving 1.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: put tangent's vertical asymptotes at x = nπ.

At those inputs cosine is 1 or −1, while sine is 0, so tangent has output 0.

✓ Instead: Put tangent's zeros at x = nπ and its vertical asymptotes at x = + nπ.

Tips and tricks

Tip: tangent's zero is halfway between consecutive asymptotes.

Strategy: step by step

1. Find the boundary lines by asking where cos x = 0. This finds the inputs that make tangent's denominator zero, so those inputs must be excluded.
2. For the central branch, use x = − $π over 2$ and x = $π over 2$ . Check cos(− $π over 2$ ) = 0 and cos( $π over 2$ ) = 0 before drawing the dashed vertical lines.
3. Plot (− $π over 4$ , −1), (0, 0), and ( $π over 4$ , 1), using the exact unit-circle values in the worked example's three-column table.
4. Draw one smooth rising branch through the points. Its left end drops without bound and its right end rises without bound as the inputs approach the boundary lines from inside the interval.
5. Copy the same branch π units right and π units left. Keep every boundary line separate from the curve.

Strategy

Strategy: sketch tangent from its denominator, points, and period

Is cos x = 0 at the input?

YesExclude the input and draw its vertical asymptote; the quotient is undefined there.

NoThe input is accepted; evaluate sine divided by cosine.

↓

Does the question ask for only one branch?

YesStay between the two specified consecutive asymptotes.

NoCopy the branch by π to cover the requested interval.

↓

Does the question ask what happens near an asymptote?

YesChoose the left or right side, then use the numerator and denominator signs.

NoGive the exact accepted values or the sketch requested.

Find and check the zeros of cosine to locate excluded inputs.
Draw two consecutive boundary lines and mark the zero halfway between them.
Use quarter-π offsets from that zero to place heights −1 and 1.
Draw one continuous rising branch, then repeat it every π.

Worked exampleThe tangent branch through (π, 0) and its vertical asymptotes

Graph the branch of y = tan x that passes through the point (π, 0). The three-column table gives exact values on this branch: tan( $3 π over 4$ ) = −1, tan(π) = 0 and tan( $5 π over 4$ ) = 1. The table lists the inputs as multiples of π, so 0.75, 1 and 1.25 stand for $3 π over 4$ , π and $5 π over 4$ . (a) Use cos x = 0 to find the two vertical asymptotes that bound this branch, and check that cos x = 0 at each one. (b) Confirm the three table values from the unit-circle values of sin x and cos x, then plot the three points. (c) Describe what the graph does as x approaches each asymptote from inside the branch, and explain why the graph never meets the line x = $3 π over 2$ . (d) Copy the branch π units left and π units right, and list every vertical asymptote in the three-branch sketch.

Three-column table for the branch of y = tan x through (π, 0). The inputs 0.75, 1 and 1.25 are multiples of π and stand for x = , π and . The outputs are tan x = −1, 0 and 1.

Solve cos x = 0. On the unit circle, cos x is 0 only at the top point (0, 1) and the bottom point (0, −1), so x = $π over 2$ + nπ for every integer n: …, − $π over 2$ , $π over 2$ , $3 π over 2$ , $5 π over 2$ , … The nearest of these on each side of π are x = $π over 2$ on the left and x = $3 π over 2$ on the right.tan x = $\frac{\sin x}{\cos x}$ , so any input that makes the denominator cos x equal 0 has no tangent value and must be excluded. The branch through (π, 0) runs from the nearest excluded input on the left of π to the nearest excluded input on the right of π.
Check both boundary inputs before drawing: cos( $π over 2$ ) = 0 (with sin( $π over 2$ ) = 1) and cos( $3 π over 2$ ) = 0 (with sin( $3 π over 2$ ) = −1). Then draw dashed vertical lines x = $π over 2$ and x = $3 π over 2$ . They are $3 π over 2$ − $π over 2$ = π apart, which is the width of one branch.A dashed line belongs only where the denominator really is 0. Do not draw one where sin x = 0. At x = π, tan(π) = 0 ÷ (−1) = 0, so that input gives a point on the graph and does not give an asymptote.
Confirm the table with exact unit-circle values. At $3 π over 4$ : cos x = − $square root of 2 over 2$ and sin x = $square root of 2 over 2$ , so tan( $3 π over 4$ ) = $square root of 2 over 2$ ÷ (− $square root of 2 over 2$ ) = −1. At π: cos x = −1 and sin x = 0, so tan(π) = 0 ÷ (−1) = 0. At $5 π over 4$ : cos x = − $square root of 2 over 2$ and sin x = − $square root of 2 over 2$ , so tan( $5 π over 4$ ) = (− $square root of 2 over 2$ ) ÷ (− $square root of 2 over 2$ ) = 1. Plot ( $3 π over 4$ , −1), (π, 0) and ( $5 π over 4$ , 1).Exact values give exact points. $3 π over 4$ is halfway between $π over 2$ and π, and $5 π over 4$ is halfway between π and $3 π over 2$ . The three points therefore mark the center of the branch and one point on each side of it.
Find the sign of tan x close to each line. Just right of $π over 2$ (Quadrant II), sin x is close to 1 and cos x is a small negative number, so the quotient is a large negative number. Just left of $3 π over 2$ (Quadrant III), sin x is close to −1 and cos x is a small negative number, so the quotient is a large positive number.Dividing a number near 1 or −1 by a number shrinking toward 0 gives outputs that grow past any bound. The signs of sin x and cos x decide whether the curve heads down or up.
Draw one smooth rising curve through the three points. As x approaches $π over 2$ from the right, the curve drops without bound beside that dashed line. As x approaches $3 π over 2$ from the left, the curve rises without bound beside the other dashed line. The curve gets ever closer to each line but never touches or crosses it.Each branch of tangent is continuous and increasing on its open interval. The inputs x = $π over 2$ and x = $3 π over 2$ are outside the domain, so no point of the graph can lie on those lines. That is the meaning of a vertical asymptote: an excluded input where the outputs grow without bound as the inputs close in on it.
Copy the branch π units left and π units right. Subtract π from each point to get (− $π over 4$ , −1), (0, 0) and ( $π over 4$ , 1) between x = − $π over 2$ and x = $π over 2$ . Add π to each point to get ( $7 π over 4$ , −1), (2π, 0) and ( $9 π over 4$ , 1) between x = $3 π over 2$ and x = $5 π over 2$ . Draw the new dashed lines x = − $π over 2$ and x = $5 π over 2$ , and keep every curve separate from every line.Tangent repeats every π, so tan(x + π) = tan x and tan(x − π) = tan x. Moving every point of one branch by π gives the neighboring branch. Moving a line x = $π over 2$ + nπ by π gives another line of the same form.

Answer

The branch lies between the vertical asymptotes x = and x = , where cos x = 0. It passes through (, −1), (π, 0) and (, 1) and rises from left to right. It drops without bound as x approaches from the right and rises without bound as x approaches from the left. It never meets x = because cos() = 0, which leaves tan x = undefined there. The three-branch sketch has vertical asymptotes x = −, x = , x = and x = . The copied branches pass through (−, −1), (0, 0), (, 1) and through (, −1), (2π, 0), (, 1).

Check Shift check: − π = −, and tan(−) = (−) ÷ = −1, which matches tan(). Also − π = , and tan() = 1, which matches tan(). Rising check: the outputs −1, 0, 1 increase as the inputs , π, increase. Asymptote check on a calculator in radian mode: ≈ 1.571 and tan(1.6) ≈ −34.23, a large negative value just right of the left line. Also ≈ 4.712 and tan(4.7) ≈ 80.71, a large positive value just left of the right line. Both match the drawn ends.

Work to write

cos x = 0 when x = $π over 2$ + nπ, so the branch through (π, 0) lies between x = $π over 2$ and x = $3 π over 2$
cos( $π over 2$ ) = 0 and cos( $3 π over 2$ ) = 0, so the dashed asymptotes are x = $π over 2$ and x = $3 π over 2$
tan( $3 π over 4$ ) = $square root of 2 over 2$ ÷ (− $square root of 2 over 2$ ) = −1
tan(π) = 0 ÷ (−1) = 0
tan( $5 π over 4$ ) = (− $square root of 2 over 2$ ) ÷ (− $square root of 2 over 2$ ) = 1
Plot ( $3 π over 4$ , −1), (π, 0) and ( $5 π over 4$ , 1), then draw one smooth rising curve through them
As x approaches $π over 2$ from the right, tan x drops without bound; as x approaches $3 π over 2$ from the left, tan x rises without bound
x = $3 π over 2$ is not in the domain because cos( $3 π over 2$ ) = 0, so the graph never touches that line
Copies: (− $π over 4$ , −1), (0, 0), ( $π over 4$ , 1) and ( $7 π over 4$ , −1), (2π, 0), ( $9 π over 4$ , 1)
Asymptotes in the sketch: x = − $π over 2$ , x = $π over 2$ , x = $3 π over 2$ , x = $5 π over 2$

The branch lies between the vertical asymptotes x = $π over 2$ and x = $3 π over 2$ , where cos x = 0. It passes through ( $3 π over 4$ , −1), (π, 0) and ( $5 π over 4$ , 1) and rises from left to right. It drops without bound as x approaches $π over 2$ from the right and rises without bound as x approaches $3 π over 2$ from the left. It never meets x = $3 π over 2$ because cos( $3 π over 2$ ) = 0, which leaves tan x = $\frac{\sin x}{\cos x}$ undefined there. The three-branch sketch has vertical asymptotes x = − $π over 2$ , x = $π over 2$ , x = $3 π over 2$ and x = $5 π over 2$ . The copied branches pass through (− $π over 4$ , −1), (0, 0), ( $π over 4$ , 1) and through ( $7 π over 4$ , −1), (2π, 0), ( $9 π over 4$ , 1).

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: plot the central zero

Find and plot tan 0. This means use the input 0 radians, find its output, and place that input and output on the graph.

The input 0 radians gives the point at the origin.

Use sin 0 = 0 and cos 0 = 1.The unit-circle point at 0 is (1, 0), with cosine first and sine second.
Divide $0 over 1$ = 0 and plot (0, 0).A zero numerator divided by a nonzero denominator is 0; graph coordinates put the input first.

Answer

tan 0 = 0.
Point: (0, 0).

Check On the turning-ray picture, the point on a horizontal ray has vertical coordinate 0 and a positive horizontal coordinate, so its tangent ratio is 0.

Rung 2Rung 2: add the quarter-π points

Add tan( $−π over 4$ ) and tan( $π over 4$ ) to the central sketch. This asks for the heights at the two inputs on either side of 0.

The highlighted radian inputs add the two heights beside the zero.

In radians, these three points lie on one rising branch.

At $π over 4$ , divide $square root of 2 over 2$ by $square root of 2 over 2$ to get 1.Sine and cosine are equal and nonzero there.
At − $π over 4$ , divide − $square root of 2 over 2$ by $square root of 2 over 2$ to get −1.Sine is negative and cosine positive, with equal magnitudes.
Plot (− $π over 4$ , −1), (0, 0), and ( $π over 4$ , 1), then connect them with a smooth rising curve inside the boundary lines.The accepted inputs and outputs in the table guide the continuous increasing branch.

Answer

tan(− $π over 4$ ) = −1.
tan( $π over 4$ ) = 1.
Central branch: rises through the three points.

Check A half-turn about the origin sends (, 1) to (, −1), which matches tangent's odd symmetry.

Rung 3Rung 3: copy one full tangent branch

Copy the central tangent branch one period to the right. This asks you to add π to its inputs, keep its outputs, and identify the new boundary lines.

The copied radian inputs retain the original output row.

This rising branch is one period, π radians, to the right.

Move the interval endpoints: − $π over 2$ + π = $π over 2$ and $π over 2$ + π = $3 π over 2$ .A horizontal copy by the period adds π to every input.
Move the key inputs: − $π over 4$ + $4 π over 4$ = $3 π over 4$ , 0 + π = π, and $π over 4$ + $4 π over 4$ = $5 π over 4$ .Common denominators let you add the π fractions while preserving their order.
Keep the respective outputs −1, 0, and 1 and draw the rising branch.Period π means tangent's outputs repeat after π.
Check cos( $π over 2$ ) = 0 and cos( $3 π over 2$ ) = 0.Plugging the endpoints into the denominator confirms that both are excluded rather than plotted points.

Answer

New interval: ( $π over 2$ , $3 π over 2$ ).
Boundary lines: x = $π over 2$ , x = $3 π over 2$ .
Key points: ( $3 π over 4$ , −1), (π, 0), ( $5 π over 4$ , 1).

Check Use unit-circle signs at and : equal magnitudes with opposite signs give −1, and equal negative signs give 1. At π, sine is 0 and cosine is −1, giving the zero at the branch's center.

Rung 4Rung 4: explain both sides near

Use the original near-asymptote table and tan(1.58) ≈ −108.6492 to explain the behavior at $π over 2$ . This asks why nearby finite outputs become large, why their signs differ across the line, and whether $π over 2$ itself has an output.

The radian table distinguishes the last left input, the excluded boundary, and a right-side input.

The curve's opposite unbounded behavior is visible on the two sides of radians.

Compare each input in the main table with $π over 2$ ≈ 1.570796: all five are smaller, and the gaps shrink from left to right.This identifies a one-sided approach from below; the inputs remain accepted radian inputs.
Read the output row from the column under 1.4 to the column under 1.57: the rounded heights rise from 5.7979 to 1255.7656.The shrinking positive cosine denominator makes the quotient increasingly large.
Compare 1.58 > 1.570796 and use quadrant II signs to explain its output ≈ −108.6492.Right of $π over 2$ , sine stays positive but cosine becomes negative, so tangent is negative; this output is rounded to four decimal places.
At $π over 2$ , substitute cos( $π over 2$ ) = 0 and reject the quotient.The denominator check excludes the boundary input. Neither the table nor the phrase unbounded gives it a real value.

Answer

Left approach: tangent grows positively without bound.
Right approach: tangent grows negatively without bound.
At $π over 2$ : tangent is undefined.
The displayed nearby outputs are finite and rounded to four decimal places.

Check Rebuild the reason without the table: sine tends toward 1, and cosine tends toward 0 with a positive sign on the left and a negative sign on the right. The supplied ten-decimal approximations sin(1.57) ≈ 0.9999996829 and cos(1.57) ≈ 0.0007963267 also give a quotient ≈ 1255.7656. You do not need to calculate a decimal-angle sine or cosine on your own to explain this growth.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: a vertical asymptote means only that a curve never touches a vertical line.

Avoiding a line does not say that nearby outputs are unbounded. A missing point at an ordinary finite height would also leave a line untouched.

✓ Instead: Define a vertical asymptote by unbounded nearby outputs on at least one side; for these parent graphs, the input on that line is excluded.

✗ Not this: Counterexample: tan() = ∞ because the table values are large.

Every displayed accepted input has a finite output. At the denominator is exactly 0, and infinity is not a real value.

✓ Instead: tan() is undefined; its nearby outputs become unbounded.

✗ Not this: Counterexample: tangent is increasing across its entire domain, including across an asymptote.

Increasing applies within each branch. A point immediately left of has a large positive output, while a larger input immediately right has a large negative output.

✓ Instead: Each tangent branch is increasing; compare inputs within the same branch when using that statement.

✗ Not this: Counterexample: every zero denominator proves a vertical asymptote. The formula has a zero denominator at x = 0.

For every nonzero x the quotient is 1: at 0.1 it is 0.1 ÷ 0.1 = 1, and at 0.01 it is 0.01 ÷ 0.01 = 1. The nearby outputs stay finite because the numerator shrinks too.

✓ Instead: A denominator zero locates an excluded input. To establish a vertical asymptote, also show unbounded nearby outputs. For tan, cot, sec and csc the numerator stays nonzero at each denominator zero.

Tips and tricks

Tip: tangent rises on every branch. Put its zero halfway between its two boundary lines.
Tip: write undefined at the exact asymptote input. Write unbounded for the nearby behavior.
Tip: rebuild the boundary locations from cos x = 0 instead of guessing from the shape.

Trap. Calling infinity the output at an asymptote. Tangent has finite outputs at every accepted input, and no output at the denominator-zero input.

Keep in mind

Infinity is not an output: tan 1.57 ≈ 1255.77 is huge but finite, and at the asymptote input itself there is no value.
Rising holds inside one branch only: tan 1.57 ≈ 1255.77, yet the larger input 1.58 gives about −108.65, across the asymptote.
Tangent's zeros sit halfway between neighboring asymptotes, at x = nπ, so never draw a wall at a zero.

Memory hookTangent climbs toward walls it never touches; the walls stand where cosine is 0, at + nπ, and each branch crosses 0 halfway between two walls.

Flash cards: say the answer out loud, then flip

What is a vertical asymptote?

A vertical line x = a near which the graph's heights grow without bound
for tan, cot, sec and csc the graph has no point on it.

What does unbounded mean?

The heights pass every ceiling, or every floor, you could name.

Name the asymptotes of tan x between 2π and 4π.

x = and x = .

Is tan 1.6 positive or negative?

Negative: 1.6 is past $π over 2$ ≈ 1.5708, in Quadrant II
tan 1.6 ≈ −34.2.

True or false: tan = ∞.

False: cos = 0, so tan is undefined. ∞ is not a number.