Graphs of cosec x, sec x, cot x: Question bank

Reminders you may need while answering:

$csc x = \frac{1}{sin x}$ , undefined at $x = nπ$ . (Denominator $= sin x$ .)
$sec x = \frac{1}{cos x}$ , undefined at $x = \frac{π}{2} + nπ$ . (Denominator $= cos x$ .)
$cot x = \frac{cos x}{sin x}$ , undefined at $x = nπ$ . (Denominator $= sin x$ .)
Here $n$ means any whole number ( $\dots, - 2, - 1, 0, 1, 2, \dots$ ).

Two pictures to hold in your head first

Before the traps, build the two mental images every question below leans on. Don't skip these — they replace the phrase "click the link to the parent page."

Picture 1 — why $1/0$ builds a wall (not just a big number)

Watch what the reciprocal $y = 1/ u$ does as the denominator $u$ shrinks toward $0$ . Feed in $u = 0.5, 0.1, 0.01, 0.001$ ; the reciprocals are $2, 10, 100, 1000$ . The smaller the bottom, the taller the output — and there is no ceiling. The curve races to $+ \infty$ from the positive side and to $- \infty$ from the negative side, so at $u = 0$ the graph has no single value: it splits into a vertical wall.

Figure — Graphs of cosec x, sec x, cot x

Picture 2 — why reciprocating FLIPS increasing into decreasing

Take positive inputs that are rising: $u = 1, 2, 3, 4$ . Their reciprocals are $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ — a falling list. As the input climbs, its upside-down twin sinks. That is the whole reason $cot x$ (the reciprocal-flavoured cousin of $tan x$ ) decreases on each branch even though $tan x$ increases.

Worked example — tan increasing vs cot decreasing (with sketch)

Worked example — signs by quadrant (the missing edge case)

True or false — justify

csc x

can equal

0

for some

x

False.

csc x = 1/ sin x

; a fraction with numerator

1

is never zero no matter how large the denominator grows, so

csc x

has no zeros at all.

sec x

and

csc x

have the exact same range.

True. Both are reciprocals of something bounded by

[- 1, 1]

, so both give

(- \infty, - 1] \cup [1, \infty)

— the open band

(- 1, 1)

is forbidden for both. See Domain and Range.

cot x

and

csc x

have the same vertical asymptotes.

True. Both die where

sin x = 0

, i.e.

x = nπ

, because

sin x

is the denominator of both. Only

sec x

is different (dies where

cos x = 0

The value

y = 0.5

appears somewhere on the graph of

sec x

False.

0.5

lies inside the forbidden band

(- 1, 1)

; since

∣ cos x ∣ \leq 1

, its reciprocal has magnitude

\geq 1

, so

sec x

can never take a value between

- 1

and

1

cot x

increases across each of its branches, just like

tan x

False. Reciprocating flips the trend (Picture 2):

cot x

decreases from

+ \infty

- \infty

on every interval

(nπ, (n + 1) π)

. See Odd and Even Functions.

sec x

is an even function.

True.

sec (- x) = 1/ cos (- x) = 1/ cos x = sec x

, because

cos

is even. Its graph is symmetric about the

y

-axis.

csc x

is an even function like

sec x

False.

csc (- x) = 1/ sin (- x) = - 1/ sin x = - csc x

, so

csc

is odd (symmetric through the origin), not even. Only

sec

is even.

The period of

cot x

2 π

, the same as

csc x

False.

cot x

repeats every

π

because its asymptotes at

x = nπ

are spaced

π

apart, so its period is ==

π

==, half that of

csc x

At every point where

sin x = 1

, the graph of

csc x

has a minimum.

True.

1/1 = 1

, and just left/right of the peak

sin x < 1

csc x > 1

— the branch turns upward, giving a minimum of the upper branch at

y = 1

sec x

touches the line

y = - 1

x = 0

False. At

x = 0

cos 0 = 1

sec 0 = + 1

;

sec x

touches

- 1

where

cos x = - 1

, i.e.

x = π, 3 π, \dots

Spot the error

csc x = 0

when

sin x = 0

, so both cross the

x

-axis at

x = nπ

Error: at

x = nπ

the denominator

sin x

0

, so

csc x

is undefined there — that is exactly where it builds a vertical asymptote (Picture 1), the opposite of crossing the axis.

sec x

blows up at

x = nπ

because that's where

cos x = 0

Error:

cos (nπ) = \pm 1 \neq = 0

sec x

blows up where

cos x = 0

, which is

x = \frac{π}{2} + nπ

— the odd multiples of

\frac{π}{2}

, not the multiples of

π

"Since

cot x = 1/ tan x

, and

tan x

increases,

cot x

must increase too."

Error: reciprocating reverses the direction of motion (Picture 2); as

tan x

climbs,

1/ tan x

falls. So

cot x

decreases on each branch.

"Solving

csc x = 0.7

: I read where the curve meets

y = 0.7

Error:

0.7 \in (- 1, 1)

is inside the forbidden band, so the

csc

curve never reaches

y = 0.7

— the equation has no solution. See Solving Trigonometric Equations.

cot x

is undefined where

cos x = 0

, since

cos x

is in it."

Error: it's the denominator

sin x

that must be nonzero. Where

cos x = 0

the numerator vanishes, giving

cot x = 0

— a zero, not an asymptote.

sec (\frac{π}{2}) = 1/ cos (\frac{π}{2}) = 1/0

, so I'll write

sec (\frac{π}{2}) = \infty

Error: you cannot assign a single value

\infty

;

sec x

has no value at

x = \frac{π}{2}

. It tends to

+ \infty

from one side and

- \infty

from the other — the function is simply undefined there.

"The graph of

csc x

shifted left by

\frac{π}{2}

gives

cot x

Error: shifting

csc x

left by

\frac{π}{2}

gives

sec x

(because

cos x = sin (x + \frac{π}{2})

cot x

is a different graph entirely — its range is all reals, unlike

sec / csc

Why questions

Why does

csc x

have a "forbidden band"

(- 1, 1)

but

cot x

does not?

Because

∣ sin x ∣ \leq 1

, its reciprocal has magnitude

\geq 1

, banning values inside

(- 1, 1)

. But

cot x = cos x / sin x

is a ratio of two varying quantities, so it can take every real value, including those inside

(- 1, 1)

Why are the asymptotes of

csc x

x = nπ

rather than at

x = \frac{π}{2} + nπ

Because a wall appears where the denominator

sin x

0

, and

sin x = 0

exactly at

x = nπ

. The points

\frac{π}{2} + nπ

are where

sin x = \pm 1

— there

csc x

merely touches

\pm 1

Why does

csc x

touch

y = 1

but never go below it on the upper branch?

sin x = 1

we get

csc x = 1

; everywhere else on that branch

0 < sin x < 1

, so

csc x > 1

. Thus

y = 1

is the lowest reachable value there — a floor it kisses but never crosses.

Why does

cot x

run from

+ \infty

down to

- \infty

inside

(0, π)

rather than the reverse?

Just right of

0

sin x \to 0^{+}

and

cos x \to 1

, so

cot x \to + \infty

. Just left of

π

sin x \to 0^{+}

but

cos x \to - 1

, so

cot x \to - \infty

. Hence it slides downward across the whole interval.

Why is

sec x

even while

cot x

is odd?

sec x

is built from

cos

, an even function, so it inherits even symmetry.

cot x = cos x / sin x

mixes even

cos

with odd

sin

; even divided by odd is odd, giving

cot (- x) = - cot x

Why can we build all three graphs "for free" from

sin

cos

tan

Because each is literally a reciprocal, and the reciprocal rule

y = 1/ u

deterministically maps zeros of

u

to walls,

\pm 1

-points of

u

to touches, and preserves the sign — so no independent memorising is needed. See Graphs of sin x, cos x, tan x.

Why does the identity

1 + cot^{2} x = csc^{2} x

guarantee

csc^{2} x \geq 1

Since

cot^{2} x \geq 0

always, adding

1

gives

csc^{2} x \geq 1

, hence

∣ csc x ∣ \geq 1

— an algebraic proof of the forbidden band. See Trigonometric Identities.

Edge cases

What is

csc x

x = 0

Undefined.

sin 0 = 0

, so

csc 0 = 1/0

has no value —

x = 0

is a vertical asymptote, not a point on the graph.

Does

sec x

have a value at

x = \frac{3 π}{2}

No.

cos (\frac{3 π}{2}) = 0

, so

sec

is undefined there and has an asymptote; approaching it,

sec x \to + \infty

on one side and

- \infty

on the other.

Can

cot x

and

csc x

ever both be defined at the same

x = nπ

No. Both share the denominator

sin x

, which is

0

x = nπ

; whenever one is undefined there, so is the other — they share the same walls.

x \to 0^{-}

, what does

csc x

approach?

sin x \to 0^{-}

(small negative), so

csc x = 1/ sin x \to - \infty

. Approaching from the other side (

0^{+}

) it instead flies to

+ \infty

What is the largest value

sin x

can take, and what does

csc x

equal there?

sin x

reaches a maximum of

1

; there

csc x = 1/1 = 1

, the smallest positive value

csc x

ever attains.

Is there any

x

where

csc x

sec x

, and

cot x

are all defined and equal?

Being equal would need

sin x = cos x

and

cos x / sin x

to match;

sin x = cos x

gives

csc x = sec x = 2

, but there

cot x = 1 \neq = 2

, so no common value exists.

At an asymptote, is the function value

+ \infty

No.

+ \infty

is not a real number and not a "value"; the function is simply undefined at the asymptote, with the graph shooting toward

\pm \infty

near it. See Vertical Asymptotes and Limits.

In which single quadrant are

csc

sec

and

cot

all positive together?

Quadrant I only

(0, \frac{π}{2})

: there

sin > 0

and

cos > 0

, so all three reciprocals inherit the positive sign.

Connections

Graphs of sin x, cos x, tan x — the parent curves you reciprocate.
Vertical Asymptotes and Limits — the " $1/0$ builds a wall" idea behind every trap.
Domain and Range — the forbidden band and undefined points.
Trigonometric Identities — $1 + cot^{2} x = csc^{2} x$ proving $∣ csc x ∣ \geq 1$ .
Solving Trigonometric Equations — why no-solution cases arise.
Odd and Even Functions — the symmetry justifications.