Reciprocal identities, quotient identities: Worked examples

This page is a worked-example gym. The parent note built the six ratios and proved the identities. Here we hit every kind of situation those identities can throw at you — every quadrant, every sign, the "flat" and "vertical" degenerate angles, a real-world problem, and an exam twist.

First we map the territory, then we walk every square of the map.

The scenario matrix

Before solving anything, let us list every distinct case class this topic contains. If a formula behaves differently in a situation, that situation gets its own row.

Cell	Case class	What is tricky here	Covered by
A	Quadrant I (all positive)	nothing — the "easy" baseline	Ex 1
B	Quadrant II ( $x < 0, y > 0$ )	$cos, tan, sec, cot$ go negative	Ex 2
C	Quadrant III ( $x < 0, y < 0$ )	$sin, cos$ negative but $tan, cot$ positive	Ex 3
D	Quadrant IV ( $x > 0, y < 0$ )	only $cos, sec$ stay positive	Ex 4
E	Degenerate: angle on an axis	some ratios become $0$ or undefined (divide by zero)	Ex 5
F	Given-one-ratio + sign clue	must pick the right sign for the partner ratio	Ex 6
G	Simplify a compound expression	"convert everything to $sin, cos$ "	Ex 7
H	Limiting behaviour ( $θ \to 9 0^{\circ}$ )	$tan, sec \to \infty$ ; watch the blow-up	Ex 8
I	Real-world word problem	translate words → ratio → answer with units	Ex 9
J	Exam twist (chained identities)	reciprocal + quotient + Pythagoras together	Ex 10

Figure — Reciprocal identities, quotient identities

We will use the standard mnemonic CAST / "All-Sin-Tan-Cos" (which ratios are positive in each quadrant), but we will derive each one, not just quote it. See Unit circle definitions for the full story of extending ratios past $9 0^{\circ}$ .

Cell A — Quadrant I baseline

Worked example Example 1 — All six ratios from a Quadrant-I point

The terminal side of $θ$ passes through $(5, 12)$ . Find all six ratios.

Forecast: every ratio here will be positive (both coordinates positive). Guess $sin$ before reading — is it $12/13$ or $5/13$ ?

Find $r$ . $r = 5^{2} + 1 2^{2} = 25 + 144 = 169 = 13$ . Why this step? $r$ is the Hypotenuse; the other five ratios all need it.
Sine & cosine. $sin θ = \frac{y}{r} = \frac{12}{13}$ , $cos θ = \frac{x}{r} = \frac{5}{13}$ . Why this step? Definitions $O / H$ and $A / H$ . Here $O = y = 12$ , $A = x = 5$ .
Tangent by quotient. $tan θ = \frac{s i n θ}{c o s θ} = \frac{12/13}{5/13} = \frac{12}{5}$ . Why this step? The quotient identity lets the $13$ 's cancel — no new triangle needed.
The three reciprocals — just flip. $csc θ = \frac{13}{12}$ , $sec θ = \frac{13}{5}$ , $cot θ = \frac{5}{12}$ . Why this step? Reciprocal identities: $csc = 1/ sin$ , $sec = 1/ cos$ , $cot = 1/ tan$ .

Verify: $sin^{2} + cos^{2} = \frac{144}{169} + \frac{25}{169} = \frac{169}{169} = 1$ ✓ (Pythagorean check). And $sin θ \cdot csc θ = \frac{12}{13} \cdot \frac{13}{12} = 1$ ✓.

Cells B, C, D — the other three quadrants

The only thing that changes across quadrants is the sign. The magnitudes come from the same $O / H$ , $A / H$ machinery. Here $r$ is always positive, so a ratio is negative exactly when its top ( $x$ or $y$ ) is negative.

Worked example Example 2 — Cell B: Quadrant II

Terminal side through $(- 4, 3)$ . Find all six ratios.

Forecast: here $x < 0, y > 0$ . Which ratios stay positive? (Only the ones built from $y$ and $r$ alone.)

$r$ . $r = (- 4)^{2} + 3^{2} = 16 + 9 = 25 = 5$ . Why this step? Squaring kills the minus sign — $r$ is a length, always $+$ .
$sin, cos$ . $sin θ = \frac{y}{r} = \frac{3}{5}$ (positive), $cos θ = \frac{x}{r} = \frac{- 4}{5}$ (negative). Why this step? $cos$ carries the sign of $x$ , which is now negative.
$tan$ by quotient. $tan θ = \frac{3/5}{- 4/5} = - \frac{3}{4}$ (negative). Why this step? positive $\div$ negative $=$ negative. Quotient identity does the sign automatically.
Reciprocals. $csc = \frac{5}{3}$ ( $+$ ), $sec = - \frac{5}{4}$ ( $-$ ), $cot = - \frac{4}{3}$ ( $-$ ). Why this step? Flipping keeps the sign. So in QII only $sin, csc$ are positive — matching "S for Sine" in CAST.

Verify: $sin^{2} + cos^{2} = \frac{9}{25} + \frac{16}{25} = 1$ ✓. Sign pattern (only $sin, csc$ positive) matches Quadrant II ✓.

Cell E — degenerate angles (where a ratio dies)

When the arrow lies exactly on an axis, one of $x, y$ is $0$ . A ratio with $0$ on top is $0$ ; a ratio with $0$ on the bottom is undefined (you cannot divide by zero).

Worked example Example 5 — Cell E: angle at

$0^{\circ}$ and $9 0^{\circ}$ Evaluate all six ratios at $θ = 0^{\circ}$ (point $(1, 0)$ ) and $θ = 9 0^{\circ}$ (point $(0, 1)$ ).

Forecast: predict which two ratios "explode" at $0^{\circ}$ . (Look for a $0$ in a denominator.)

At $0^{\circ}$ , $(x, y) = (1, 0)$ , $r = 1$ :

$sin 0^{\circ} = \frac{y}{r} = \frac{0}{1} = 0$ , $cos 0^{\circ} = \frac{1}{1} = 1$ .
$tan 0^{\circ} = \frac{s i n}{c o s} = \frac{0}{1} = 0$ .
$csc 0^{\circ} = \frac{1}{s i n} = \frac{1}{0} =$ undefined; $cot 0^{\circ} = \frac{c o s}{s i n} = \frac{1}{0} =$ undefined. Why this step? Both $csc, cot$ have $sin = 0$ in the denominator → division by zero → no value exists.
$sec 0^{\circ} = \frac{1}{c o s} = \frac{1}{1} = 1$ (fine).

At $9 0^{\circ}$ , $(x, y) = (0, 1)$ , $r = 1$ : 5. $sin 9 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ . 6. $tan 9 0^{\circ} = \frac{1}{0} =$ undefined; $sec 9 0^{\circ} = \frac{1}{0} =$ undefined. Why this step? Now $cos = 0$ sits underneath → $tan, sec$ blow up. 7. $csc 9 0^{\circ} = \frac{1}{1} = 1$ , $cot 9 0^{\circ} = \frac{c o s}{s i n} = \frac{0}{1} = 0$ .

Verify: at every angle, the ratio-partner of an undefined value has value $0$ (since $undefined = 1/0$ requires its partner $= 0$ ). E.g. $csc 0^{\circ}$ undefined ↔ $sin 0^{\circ} = 0$ ✓.

Cell F — given one ratio + a sign clue

Cell G — simplify a compound expression

Cell H — limiting behaviour

Worked example Example 8 — Cell H: what happens as

$θ \to 9 0^{\circ}$ ? Describe the behaviour of $tan θ$ and $sec θ$ as $θ$ climbs from $0^{\circ}$ toward $9 0^{\circ}$ .

Forecast: $cos θ \to 0$ from the positive side. What does $1/ cos θ$ do? Guess a number: $sec 89. 9^{\circ}$ — big or small?

Track $cos θ$ . As $θ \to 9 0^{\circ}$ , $cos θ \to 0^{+}$ (small, positive). Why this step? $sec$ and $tan$ both have $cos$ in the denominator, so $cos$ 's shrinking controls them.
$sec θ = \frac{1}{c o s θ}$ grows without bound: $sec 6 0^{\circ} = 2$ , $sec 8 0^{\circ} \approx 5.76$ , $sec 8 9^{\circ} \approx 57.3$ . Why this step? Dividing $1$ by an ever-smaller positive number gives an ever-larger result.
$tan θ = \frac{s i n θ}{c o s θ}$ does the same: $tan 8 9^{\circ} \approx 57.3$ (since $sin \to 1$ , $cos \to 0^{+}$ ).
At exactly $9 0^{\circ}$ : both are undefined (Cell E) — the " $\to \infty$ " is a trend, not a reached value.

Verify: $sec 8 9^{\circ} = \frac{1}{c o s 8 9 ^{\circ}} = \frac{1}{0.017452} \approx 57.30$ ✓, and $tan 8 9^{\circ} \approx 57.29$ ✓ — both large, confirming the blow-up.

Cell I — real-world word problem

Worked example Example 9 — Cell I: a ladder and a reciprocal

A ladder leans on a wall making angle $θ$ with the ground. Its foot is $A = 2 m$ from the wall and it reaches $O = 4 m$ up. A safety code is written in terms of $csc θ$ (ratio of ladder length to height). Find the ladder length $H$ and $csc θ$ .

Forecast: $csc = H / O$ — so once we know $H$ , this is one division. Guess whether $csc θ > 1$ (it always is, for $0 < θ < 9 0^{\circ}$ ).

Find $H$ (ladder length). $H = A^{2} + O^{2} = 2^{2} + 4^{2} = 20 = 25 \approx 4.47 m$ . Why this step? $H$ is the hypotenuse — the physical ladder — via Pythagoras.
$sin θ = \frac{O}{H} = \frac{4}{2 5} = \frac{2}{5}$ . Why this step? height-over-ladder is exactly $sin$ .
$csc θ = \frac{1}{s i n θ} = \frac{5}{2} \approx 1.118$ . Why this step? Reciprocal identity — the safety ratio is just $1/ sin$ . Units cancel (it is a pure ratio, ladder-metres over height-metres).

Verify: $csc θ = \frac{H}{O} = \frac{2 5}{4} = \frac{5}{2} \approx 1.118$ ✓ (matches step 3, computed the direct way). Ratio $> 1$ as forecast ✓.

Cell J — exam-style chained twist

Worked example Example 10 — Cell J: reciprocal + quotient + Pythagoras

If $sec θ = \frac{13}{5}$ and $θ$ is acute, show that $csc θ + cot θ$ simplifies to a single fraction, and evaluate it.

Forecast: we know $sec$ , so $cos$ comes for free; then Pythagoras gives $sin$ ; then two reciprocal/quotient flips. Guess: is the answer $> 1$ ?

$cos θ = \frac{1}{s e c θ} = \frac{5}{13}$ . Why this step? Reciprocal identity — flip $sec$ .
$sin θ = 1 - cos^{2} θ = 1 - \frac{25}{169} = \frac{144}{169} = \frac{12}{13}$ (positive, acute). Why this step? Pythagorean identities supply the missing partner; acute ⟹ $sin > 0$ .
$csc θ = \frac{1}{s i n θ} = \frac{13}{12}$ . Why this step? Reciprocal flip.
$cot θ = \frac{c o s θ}{s i n θ} = \frac{5/13}{12/13} = \frac{5}{12}$ . Why this step? Quotient identity; the $13$ 's cancel.
Add. $csc θ + cot θ = \frac{13}{12} + \frac{5}{12} = \frac{18}{12} = \frac{3}{2}$ . Why this step? Common denominator $12$ ; simplify.

Verify: $\frac{3}{2} = 1.5 > 1$ as forecast ✓. Cross-check: $csc - cot = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3}$ , and $(csc + cot) (csc - cot) = csc^{2} - cot^{2} = 1$ (a Pythagorean identity); indeed $\frac{3}{2} \cdot \frac{2}{3} = 1$ ✓.

Recall Quick self-test across the matrix

In which quadrant is only $tan, cot$ positive? ::: Quadrant III $csc 0^{\circ} = ?$ ::: undefined (since $sin 0^{\circ} = 0$ ) If $tan θ < 0$ , which two quadrants are possible? ::: II and IV As $θ \to 9 0^{\circ}$ , what happens to $sec θ$ ? ::: it grows without bound (undefined at exactly $9 0^{\circ}$ )

Connections

Parent: Reciprocal & Quotient Identities — the definitions and proofs behind every example here.
Unit circle definitions — the source of quadrant signs used in Cells B–E.
Pythagorean identities — supplies missing ratios in Ex 2–10 and the sanity checks.
Right triangle trigonometry (SOH-CAH-TOA) — where $O / H$ , $A / H$ , $O / A$ come from.
Simplifying trigonometric expressions — the "convert to $sin, cos$ " trick of Cell G.
Inverse trigonometric functions — recall $csc \neq = sin^{- 1}$ .

Worked examples — Reciprocal identities, quotient identities