Reciprocal identities — cosec, sec, cot in terms of sin, cos, tan: Question bank

Before the questions, one reminder that most traps below feed on:

True or false — justify

True or false:

csc θ

and

sin^{- 1} θ

are the same thing.

False.

csc θ = \frac{1}{s i n θ}

is a reciprocal (a number you flip), while

sin^{- 1} θ = arcsin θ

is an angle whose sine is

θ

; e.g.

csc 3 0^{\circ} = 2

but

arcsin 30

is undefined since arcsin only accepts inputs in

[- 1, 1]

True or false: For any angle,

csc θ \times sin θ = 1

Only where

sin θ \neq = 0

. Since

csc θ = \frac{1}{s i n θ}

, the product is

\frac{s i n θ}{s i n θ} = 1

— but at

θ = 0^{\circ}, 18 0^{\circ}, \dots

the

csc

is undefined, so no product exists there.

True or false:

sec θ

is always at least

1

in size.

True in magnitude:

∣ cos θ ∣ \leq 1

, so

∣ sec θ ∣ = \frac{1}{∣ c o s θ ∣} \geq 1

. Secant can never land strictly between

- 1

and

1

; the same argument makes

∣ csc θ ∣ \geq 1

always.

True or false:

cot θ = \frac{c o s θ}{s i n θ}

and

cot θ = \frac{1}{t a n θ}

are equal for every angle where either is defined.

Not quite — they agree wherever both are defined, but

\frac{c o s θ}{s i n θ}

is still meaningful at

θ = 9 0^{\circ}

(giving

0

), whereas

\frac{1}{t a n θ}

chokes because

tan 9 0^{\circ}

is itself undefined. The

cos / sin

form has the cleaner domain.

True or false: As

θ

grows from

0^{\circ}

toward

9 0^{\circ}

sec θ

increases.

True.

cos θ

shrinks from

1

down toward

0

, so its reciprocal

sec θ

climbs from

1

upward without bound — a smaller positive denominator makes a bigger quotient.

True or false: If

sin θ

is negative then

csc θ

is negative.

True. Flipping a fraction keeps its sign:

\frac{1}{( negative )}

is negative. So

csc

sec

cot

each carry the same sign as

sin

cos

tan

respectively.

True or false:

sec θ = csc (9 0^{\circ} - θ)

True — this is a complementary-angle fact. Since

cos θ = sin (9 0^{\circ} - θ)

, flipping both sides gives

sec θ = csc (9 0^{\circ} - θ)

; the "co" in cosecant literally records this partnership.

True or false: The identity

1 + cot^{2} θ = csc^{2} θ

is a brand-new fact independent of

sin^{2} θ + cos^{2} θ = 1

False. It is the Pythagorean identity divided through by

sin^{2} θ

; nothing new enters, we just renamed

\frac{c o s ^{2}}{s i n ^{2}}

cot^{2}

and

\frac{1}{s i n ^{2}}

csc^{2}

Spot the error

Find the flaw: "

csc 0^{\circ} = \frac{1}{s i n 0 ^{\circ}} = \frac{1}{0} = \infty

, so

csc 0^{\circ} = \infty

\frac{1}{0}

is undefined, not a number called

\infty

csc 0^{\circ}

does not exist; the graph has a vertical asymptote there — values grow arbitrarily large near

0^{\circ}

but never reach a value at it.

Find the flaw: "Since

sec θ = \frac{1}{c o s θ}

, we have

sec 9 0^{\circ} = \frac{1}{c o s 9 0 ^{\circ}} = \frac{1}{0} = 0

Two errors:

\frac{1}{0}

isn't

0

, and

sec 9 0^{\circ}

is simply undefined. Dividing

1

by the tiny numbers near

cos 9 0^{\circ} = 0

blows the reciprocal up, not down to zero.

Find the flaw: "

cot θ = \frac{c o s θ}{s i n θ}

, so

cot θ

is undefined exactly where

cos θ = 0

Wrong denominator. A fraction dies when its bottom is zero, so

cot

is undefined where

sin θ = 0

(at

0^{\circ}, 18 0^{\circ}, \dots

). At

cos θ = 0

the top is zero, which just makes

cot = 0

, perfectly defined.

Find the flaw: "To simplify

\frac{s e c θ}{c s c θ}

, cancel the

sec

and

csc

to get

1

You can't cancel unlike names. Rewrite first:

\frac{1/ c o s θ}{1/ s i n θ} = \frac{s i n θ}{c o s θ} = tan θ

, not

1

Find the flaw: "

csc θ = 2

means

sin θ = 2

Backwards.

csc θ = 2

means

\frac{1}{s i n θ} = 2

, so

sin θ = \frac{1}{2}

. A sine of

2

is impossible anyway, since

∣ sin θ ∣ \leq 1

Find the flaw: "

(sin θ)^{- 1} = csc θ

and

sin^{- 1} θ = csc θ

— same thing written two ways."

Only the first is right.

(sin θ)^{- 1}

raises the value

sin θ

- 1

(a reciprocal

= csc θ

), but

sin^{- 1} θ

is the inverse function arcsin. The parentheses change everything.

Why questions

Why do we even bother naming

csc, sec, cot

instead of just writing

\frac{1}{s i n}

etc.?

Because dividing by a trig function shows up constantly (wave physics, integrals, derivatives); a single symbol keeps expressions readable and lets us write clean identities like

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

Why is

cot θ = \frac{c o s θ}{s i n θ}

often more useful than

\frac{1}{t a n θ}

It shares the language of

sin

and

cos

, so terms cancel directly instead of building nested fractions like

\frac{1}{s i n / c o s}

. In simplifications everything is already in

sin / cos

, so

cot

slots in cleanly.

Why does

sec θ

have the same undefined points as

tan θ

Both are secretly divided by

cos θ

sec θ = \frac{1}{c o s θ}

and

tan θ = \frac{s i n θ}{c o s θ}

. Wherever

cos θ = 0

(at

9 0^{\circ}, 27 0^{\circ}, \dots

) both blow up together.

Why can

sec θ

never equal, say,

0.5

It's a reciprocal of

cos θ

, and

∣ cos θ ∣ \leq 1

forces

∣ sec θ ∣ \geq 1

. The whole band

(- 1, 1)

is a forbidden zone for

sec

and

csc

values.

Why does solving

csc θ = 1

give the same angles as

sin θ = 1

Flipping a fraction to itself needs the value to be its own reciprocal — and

1 = \frac{1}{1}

. So

csc θ = 1

exactly when

sin θ = 1

, useful when reducing a basic trig equation with reciprocals to a sine/cosine one.

Edge cases

θ = 9 0^{\circ}

: which of

csc, sec, cot

exist, and what are they?

Here

sin = 1, cos = 0, tan

undefined. So

csc 9 0^{\circ} = 1

(defined),

cot 9 0^{\circ} = \frac{0}{1} = 0

(defined), but

sec 9 0^{\circ}

is undefined because

cos 9 0^{\circ} = 0

θ = 0^{\circ}

: which reciprocal functions are defined?

sin 0^{\circ} = 0

and

tan 0^{\circ} = 0

, so

csc 0^{\circ}

and

cot 0^{\circ}

are both undefined. Only

sec 0^{\circ} = \frac{1}{c o s 0 ^{\circ}} = \frac{1}{1} = 1

survives.

What happens to

csc θ

θ

shrinks toward

0^{\circ}

from the positive side versus the negative side?

From the positive side

sin θ \to 0^{+}

csc θ \to + \infty

; from the negative side

sin θ \to 0^{-}

csc θ \to - \infty

. The two-sided limits disagree, confirming a genuine asymptote rather than a value.

Is there any angle where

csc θ = sec θ

Yes: this needs

sin θ = cos θ

, i.e.

θ = 4 5^{\circ}

(and

22 5^{\circ}

), where both equal

2

(or

- 2

). It's the same balance point that makes

tan θ = 1

Can

cot θ

and

tan θ

ever both be zero at the same angle?

Never. Where

tan θ = 0

(so

sin θ = 0

cot θ = \frac{c o s θ}{s i n θ}

divides by zero and is undefined. One being zero forces the other to be undefined — they're reciprocals.

Recall One-line self-test

Cover and answer: why does the range $(- 1, 1)$ never contain a value of $sec$ or $csc$ ? Because they are reciprocals of quantities capped at magnitude $1$ , and flipping something $\leq 1$ in size gives something $\geq 1$ in size.