Reciprocal identities, quotient identities: Question bank

Before we start, one shared picture to keep in your head:

Figure — Reciprocal identities, quotient identities

Figure s01 — Alt-text / caption: A point labelled

(x, y)

sits on a circle of radius

r

centred at the origin. The magenta arrow from the origin to

(x, y)

is the radius

r

. The orange horizontal leg along the

x

-axis is

x = r cos θ

; the violet vertical leg is

y = r sin θ

. The navy arc at the origin marks the angle

θ

from the positive

x

-axis. The box lists all six ratios read off these three lengths:

sin θ = y / r

cos θ = x / r

tan θ = y / x

, and their reciprocals

csc θ = r / y

sec θ = r / x

cot θ = x / y

The point $(x, y)$ sits on a circle of radius $r$ . From the three lengths $x$ , $y$ , $r$ you read off all six ratios: $sin θ = y / r$ , $cos θ = x / r$ , $tan θ = y / x$ , and by flipping them $csc θ = r / y$ , $sec θ = r / x$ , $cot θ = x / y$ . Every trap below is really a statement about $x$ , $y$ , or $r$ — so when in doubt, return to this diagram.

True or false — justify

A "reciprocal" of a number means flipping it, so

csc θ = (sin θ)^{- 1}

; the "-1" here is a power.

True.

(sin θ)^{- 1} = 1/ sin θ = csc θ

— the exponent

- 1

genuinely means "reciprocal of the value." The confusion only starts when the

- 1

is glued to the function name instead (see next).

sin^{- 1} θ

and

csc θ

are two names for the same thing.

False.

sin^{- 1}

is the inverse function (arcsin) that undoes sine and returns an angle;

csc θ = (sin θ)^{- 1}

is the reciprocal of a value. See Inverse trigonometric functions — they answer completely different questions.

tan θ = sin θ / cos θ

holds for absolutely every angle

θ

False. It fails wherever

cos θ = 0

(at

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

), because you cannot divide by zero — there

tan θ

is undefined, matching the vertical-line picture on the unit circle.

cot θ = 1/ tan θ

is valid even when

tan θ = 0

False. When

tan θ = 0

(at

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

sin θ = 0

too, so both

1/ tan θ

and

cos θ / sin θ

divide by zero —

cot θ

simply does not exist at those angles, by any route.

sin θ csc θ = 1

is true for every angle without exception.

False. It is true only where both are defined; at

θ = 0^{\circ}

sin θ = 0

csc θ

doesn't exist and the product

0 \cdot (undefined)

is meaningless, not

1

Since

cos

and

sec

are reciprocals, if

cos θ

is small then

sec θ

is large.

True.

sec θ = 1/ cos θ

, so as

cos θ \to 0

sec θ \to \pm \infty

. This is why

sec

(and

csc

) have values that shoot off to infinity but never sit between

- 1

and

1

sec θ

can equal

0.5

False.

sec θ = 1/ cos θ

and

∣ cos θ ∣ \leq 1

, so

∣ sec θ ∣ \geq 1

always. No secant value ever lands strictly inside

(- 1, 1)

The identity

tan θ cot θ = 1

means

tan

and

cot

are always both positive.

False. It only means they have the same sign (product

+ 1

), so their reciprocity holds; both can be negative together, e.g. in Quadrant II where

tan < 0

and

cot < 0

and

tan cot = 1

still.

Spot the error

cot θ = 1/ tan θ = sin θ / cos θ

." Find the slip.

The last step is upside down.

1/ tan θ = 1/ (sin θ / cos θ) = cos θ / sin θ

. Anchor:

tan

has

sin

on top, so its reciprocal

cot

has

cos

on top.

sec θ = 1/ sin θ

because both start with the letter s."

The pairing is a cross-swap, not a letter match.

sec \leftrightarrow cos

and

csc \leftrightarrow sin

. The correct statement is

sec θ = 1/ cos θ

"Given

sin θ = 2/3

, then

csc θ = sin^{- 1} (2/3) \approx 41. 8^{\circ}

Two errors fused:

csc

is a reciprocal, not arcsin, and it returns a number, not an angle. Correctly

csc θ = 1/ (2/3) = 3/2

\frac{tan θ}{sec θ} = \frac{sin / cos}{1/ cos} = \frac{sin}{cos} \cdot \frac{1}{cos} = \frac{sin}{cos ^{2}}

The reciprocal was mis-flipped. Dividing by

1/ cos

means multiplying by $cos$ , not by

1/ cos

: result is

sin θ

, not

sin / cos^{2}

"To simplify

cot θ \cdot sin θ

I turn it into

\frac{cos θ}{sin θ} \cdot sin θ = cos θ

— valid for all

θ

The algebra is right, but the cancellation of

sin θ

silently assumes

sin θ \neq = 0

. At

θ = 0^{\circ}

the original

cot θ

is undefined, so the tidy result

cos θ

only holds where

sin θ \neq = 0

cos θ = 1 - sin^{2} θ

, so from

sin θ

I can always get

cos θ

The square root drops the sign. It gives

∣ cos θ ∣

; you must choose

+

-

using the quadrant of

θ

. In Quadrants II and III,

cos θ

is negative. See Pythagorean identities.

Why questions

Why do we only ever need to memorise

sin

and

cos

Because

tan = sin / cos

(quotient) and

csc, sec, cot

are reciprocals of

sin, cos, tan

. From two ratios the other four are generated — no independent memory required.

Why is "convert everything to

sin

and

cos

" the universal simplification move?

Because both quotient and reciprocal identities express every ratio through

sin

and

cos

, so rewriting collapses six different symbols into one common language where factors cancel cleanly. This lives in Simplifying trigonometric expressions.

Why does

tan θ

repeat every

18 0^{\circ}

but

sin θ

repeats every

36 0^{\circ}

tan = y / x

; flipping the point to

(- x, - y)

18 0^{\circ}

turn) flips both signs, leaving the ratio unchanged.

sin = y / r

depends on

y

alone, which does change sign under that turn.

Why can

csc

and

sec

never take values between

- 1

and

1

They are reciprocals of

sin

and

cos

, whose magnitudes never exceed

1

. Flipping a number of size

\leq 1

gives a number of size

\geq 1

Why is the identity

tan θ = sin θ / cos θ

preferable to the triangle definition

O / A

for angles past

9 0^{\circ}

Right triangles only house acute angles, but

sin

and

cos

are defined for all angles via Unit circle definitions. The quotient identity carries

tan

to every angle where

cos \neq = 0

, beyond what SOH-CAH-TOA can reach.

Why does the SOH-CAH-TOA triangle only justify these identities for

0^{\circ} < θ < 9 0^{\circ}

Because a triangle's sides

O, A, H

are positive lengths and its angle is acute; the identities extend to other quadrants only after signed coordinates from the unit circle take over. See Right triangle trigonometry (SOH-CAH-TOA).

Edge cases

θ = 9 0^{\circ}

, which of the six ratios exist?

sin = 1

csc = 1

cos = 0

, and

cot = cos / sin = 0

all exist. But

tan

and

sec

divide by

cos = 0

, so both are undefined.

θ = 0^{\circ}

, which ratios blow up?

sin = 0

and

cos = 1

, so

csc = 1/0

and

cot = cos / sin = 1/0

are undefined;

tan = 0

sec = 1

, and

sin, cos

are all fine.

If a point on the terminal side is the origin

(0, 0)

, what are the trig ratios?

All undefined:

r = 0

, so every ratio divides by zero. The angle isn't determined by a zero-length "vector," so no ratio can be assigned.

In Quadrant III, both

sin θ

and

cos θ

are negative — what happens to

tan θ

and

cot θ

Negative

\div

negative

=

positive, so

tan θ > 0

and

cot θ > 0

in Quadrant III, even though

sin

and

cos

are each negative there.

Can

sin θ

and

csc θ

have opposite signs?

No.

csc θ = 1/ sin θ

, and reciprocating never changes sign, so they always share the same sign (positive together or negative together).

θ

creeps toward

9 0^{\circ}

from below, what does

tan θ

do, and why?

It grows without bound toward

+ \infty

, because

tan = sin / cos

and

cos θ \to 0^{+}

while

sin θ \to 1

— dividing a fixed positive number by a shrinking positive one explodes.

cos θ = 0

, why is it wrong to "simplify"

tan θ \cdot cos θ

sin θ

at that angle?

Writing

tan θ cos θ = (sin / cos) cos = sin

cancels

cos

, but

tan θ

is itself undefined when

cos θ = 0

, so the left side never existed to be simplified there.

Connections

Parent: Reciprocal & Quotient Identities
Pythagorean identities — needed to recover $cos$ from $sin$ with the right sign.
Unit circle definitions — the reason identities survive past $9 0^{\circ}$ .
Inverse trigonometric functions — the $sin^{- 1}$ vs $csc$ trap headquarters.
Simplifying trigonometric expressions — where "convert to sin & cos" pays off.
Right triangle trigonometry (SOH-CAH-TOA) — the acute-only origin of the six ratios.