Reciprocal identities — cosec, sec, cot in terms of sin, cos, tan: Worked examples

This page is the "no surprises" companion to the parent note. There we defined the three reciprocal functions. Here we fight through every case they can throw at you — every sign, every quadrant, the values where they explode, a real-world problem, and an exam-style trick. When you finish, no exam question should feel unfamiliar.

Before we start, three plain-word reminders (never assume — always rebuild):

Two ideas we lean on repeatedly:

Sign of a flip = sign of the original. If $sin θ$ is negative, then $1/ sin θ$ is negative too — dividing $1$ by a negative number gives a negative number. So $csc$ copies the sign of $sin$ , $sec$ copies $cos$ , $cot$ copies $tan$ .
A flip explodes where the original hits zero. $1/0$ is undefined — you cannot share one thing among zero people. So each reciprocal has "forbidden" angles.

Definition The forbidden angles and how often they repeat (in radians)

A reciprocal is undefined exactly where its parent equals zero. Let $k$ stand for any whole number ( $\dots, - 2, - 1, 0, 1, 2, \dots$ ):

$csc θ$ undefined where $sin θ = 0$ , i.e. at $θ = k π$ (that is $0, \pm 180°, \pm 360°, \dots$ ).
$cot θ = \frac{cos θ}{sin θ}$ also undefined at $θ = k π$ (same zeros of $sin$ in the denominator).
$sec θ$ undefined where $cos θ = 0$ , i.e. at $θ = \frac{π}{2} + k π$ (that is $90°, 270°, \dots$ ).

How often each flip repeats (its period):

$csc θ$ repeats every $2 π$ ( $360°$ ), because $sin$ does. So its danger points $θ = k π$ recur, but the shape of the curve returns only after a full $2 π$ turn.
$sec θ$ likewise has period $2 π$ ( $360°$ ), because $cos$ does.
$cot θ$ has the shorter period $π$ ( $180°$ ): $tan$ and $cot$ repeat twice as fast as $sin$ and $cos$ , so $cot$ 's whole graph comes back every $180°$ .

Knowing one period of each function therefore tells you its entire infinite graph — one "danger list" per function is enough forever.

The scenario matrix

Every reciprocal-identity problem is one (or a blend) of these cells. The worked examples below are labelled by cell so you can see the whole space is covered.

Cell	What makes it different	Danger to watch
A. Quadrant I	all of $sin, cos, tan$ positive	none — the "easy" case
B. Quadrant II	$sin > 0$ , $cos < 0$ , $tan < 0$	sign of $sec, cot$
C. Quadrant III	$sin < 0$ , $cos < 0$ , $tan > 0$	sign of $csc, sec$
D. Quadrant IV	$sin < 0$ , $cos > 0$ , $tan < 0$	sign of $csc, cot$
E. Degenerate / zero	$sin θ = 0$ or $cos θ = 0$	reciprocal is undefined
F. Limiting behaviour	$θ$ approaches a forbidden angle	value grows without bound
G. Word problem	physical ramp / distance	keep units, pick the right flip
H. Exam twist	prove/simplify an identity	use $cot = cos / sin$ , Pythagorean link

We fill every cell across Examples 1–10.

The quadrant sign chart (build it once, use it forever)

Where in a full turn is each basic function positive? Look at the coordinate plane: a point on the unit circle has coordinates $(cos θ, sin θ)$ . So $cos$ is the $x$ -coordinate and $sin$ is the $y$ -coordinate.

Figure — Reciprocal identities — cosec, sec, cot in terms of sin, cos, tan

Reading the figure: the white circle is the unit circle (radius $1$ ). The pink arrow points to a sample angle's location; its horizontal shadow onto the $x$ -axis is $cos θ$ , its vertical shadow onto the $y$ -axis is $sin θ$ . Each quadrant is labelled with the signs of all six functions — the three parents $sin, cos, tan$ and their three flips $csc, sec, cot$ . Trace the arrow anticlockwise through all four quadrants and watch the signs flip as it crosses each axis:

Right half of the picture ( $x > 0$ ): $cos > 0 \Rightarrow sec > 0$ .
Top half ( $y > 0$ ): $sin > 0 \Rightarrow csc > 0$ .
$tan = sin / cos$ (and its flip $cot$ ) is positive where $x, y$ share a sign — Quadrants I and III.

Worked examples

Example 1 — Cell A (Quadrant I, clean numbers)

Example 2 — Cell A (Quadrant I, Pythagorean link)

Worked example

$sin θ = \frac{5}{13}$ with $θ$ acute. Find $csc θ$ , $sec θ$ , $cot θ$ . Forecast: $5$ – $12$ – $13$ is a famous right triangle. Which side is missing here?

Step 1. $csc θ = \frac{1}{5/13} = \frac{13}{5}$ . Why this step? Direct flip of the given $sin$ — the fastest reciprocal to write down, so we bank it first.

Step 2. Find $cos θ$ using the Pythagorean identity $sin^{2} θ + cos^{2} θ = 1$ : $cos^{2} θ = 1 - \frac{25}{169} = \frac{144}{169}, cos θ = \frac{12}{13} .$ Why this step? We need $cos$ (and hence $tan$ ) before we can flip to $sec$ and $cot$ . We take the positive root because Quadrant I ( $θ$ acute) makes $cos > 0$ .

Step 3. Flip and combine: $sec θ = \frac{13}{12}, cot θ = \frac{c o s θ}{s i n θ} = \frac{12/13}{5/13} = \frac{12}{5} .$ Why this step? Using $cot = cos / sin$ (not $1/ tan$ ) skips a messy nested fraction.

Verify: $csc^{2} θ = \frac{169}{25}$ and $1 + cot^{2} θ = 1 + \frac{144}{25} = \frac{169}{25}$ . They match, confirming $1 + cot^{2} θ = csc^{2} θ$ ✓.

Example 3 — Cell B (Quadrant II)

Worked example

$θ$ is in Quadrant II and $sin θ = \frac{4}{5}$ . Find $csc θ$ , $sec θ$ , $cot θ$ . Forecast: In Quadrant II, which of the three flips come out negative?

Step 1. $csc θ = \frac{1}{4/5} = \frac{5}{4}$ (positive — matches $sin > 0$ ). Why this step? $csc$ is the direct flip of the given $sin$ , and $sin > 0$ throughout Quadrant II, so no sign trap here — we secure the safe value first.

Step 2. $cos^{2} θ = 1 - \frac{16}{25} = \frac{9}{25}$ , so $∣ cos θ ∣ = \frac{3}{5}$ . In Quadrant II $cos < 0$ , so $cos θ = - \frac{3}{5} .$ Why this step? From the sign chart, the left half of the plane ( $x < 0$ ) forces $cos$ negative. We must choose the negative root — this is exactly where naive students slip.

Step 3. Flip: $sec θ = \frac{1}{- 3/5} = - \frac{5}{3}, cot θ = \frac{c o s θ}{s i n θ} = \frac{- 3/5}{4/5} = - \frac{3}{4} .$ Why this step? $sec$ inherits the minus sign from $cos$ ; $cot$ inherits it from $tan$ (which is negative in QII).

Verify: $sec^{2} θ - tan^{2} θ$ should be $1$ . First compute $tan θ = \frac{sin θ}{cos θ} = \frac{4/5}{- 3/5} = - \frac{4}{3}$ , so $tan^{2} θ = (- \frac{4}{3})^{2} = \frac{16}{9}$ . Then $sec^{2} θ - tan^{2} θ = \frac{25}{9} - \frac{16}{9} = \frac{9}{9} = 1$ ✓.

Example 4 — Cell C (Quadrant III)

Worked example

$θ$ in Quadrant III with $tan θ = \frac{3}{4}$ . Find $cot θ$ , $sec θ$ , $csc θ$ . Forecast: $tan$ is positive in QIII (two negatives divide to a positive). So which flips end up negative here?

Step 1. $cot θ = \frac{1}{3/4} = \frac{4}{3}$ (positive — matches $tan > 0$ ). Why this step? $cot$ is the direct flip of the given $tan$ ; both share a sign, so this value is safe to write immediately.

Step 2. Build the reference triangle: opposite $= 3$ , adjacent $= 4$ , hypotenuse $= 3^{2} + 4^{2} = 5$ . Attach signs for QIII where both $x$ and $y$ are negative, so $sin < 0$ and $cos < 0$ : $sin θ = - \frac{3}{5}, cos θ = - \frac{4}{5} .$ Why this step? The triangle gives magnitudes; the quadrant assigns signs. Never skip the sign step.

Step 3. Flip: $csc θ = - \frac{5}{3}, sec θ = - \frac{5}{4} .$ Why this step? Both parents ( $sin, cos$ ) are negative, so both flips are negative.

Verify: $sin^{2} θ + cos^{2} θ = \frac{9}{25} + \frac{16}{25} = 1$ ✓, and $cot θ \cdot tan θ = \frac{4}{3} \cdot \frac{3}{4} = 1$ ✓ (a reciprocal times its own function is always $1$ ).

Example 5 — Cell D (Quadrant IV)

Worked example

$θ$ in Quadrant IV with $cos θ = \frac{12}{13}$ . Find $sec θ$ , $csc θ$ , $cot θ$ . Forecast: $cos$ is positive in QIV (right half). What sign will $csc$ and $cot$ carry?

Step 1. $sec θ = \frac{13}{12}$ (positive — matches $cos > 0$ ). Why this step? $sec$ is the direct flip of the given $cos$ ; $cos > 0$ throughout Quadrant IV, so this reciprocal has no sign trap — we lock it in first.

Step 2. $sin^{2} θ = 1 - \frac{144}{169} = \frac{25}{169}$ , so $∣ sin θ ∣ = \frac{5}{13}$ . In QIV, $y < 0$ so $sin < 0$ : $sin θ = - \frac{5}{13} .$ Why this step? The bottom half of the plane ( $y < 0$ ) forces $sin$ negative, so we pick the negative root of $sin^{2} θ$ .

Step 3. Flip: $csc θ = - \frac{13}{5}, cot θ = \frac{c o s θ}{s i n θ} = \frac{12/13}{- 5/13} = - \frac{12}{5} .$ Why this step? $csc$ copies $sin$ 's minus; $cot$ copies $tan$ 's minus (negative over positive).

Verify: $1 + cot^{2} θ = 1 + \frac{144}{25} = \frac{169}{25}$ and $csc^{2} θ = \frac{169}{25}$ — equal ✓.

Example 6 — Cell E (degenerate / undefined inputs)

Worked example Evaluate (if possible)

$csc 0°$ , $sec 90°$ , $cot 180°$ . Forecast: Which of these three is a valid finite number?

Step 1. $sin 0° = 0 \Rightarrow csc 0° = \frac{1}{0}$ → undefined. Not infinity — just undefined. Why this step? Dividing by zero has no answer at all; there is no number you can multiply by $0$ to get $1$ . And $0° = 0 π$ sits in the forbidden family $θ = k π$ .

Step 2. $cos 90° = 0 \Rightarrow sec 90° = \frac{1}{0}$ → undefined. Why this step? $90° = \frac{π}{2} + 0 π$ is in $sec$ 's forbidden family $θ = \frac{π}{2} + k π$ .

Step 3. $cot 180° = \frac{cos 180°}{sin 180°} = \frac{- 1}{0}$ → undefined. Why this step? $cot$ blows up wherever $sin = 0$ , which is the family $θ = k π$ ; $180° = 1 \cdot π$ is in it.

Verify (domain rule): Before flipping, check the parent is non-zero. $sin 0° = sin 180° = 0$ and $cos 90° = 0$ — all forbidden. The lesson: always test the denominator first.

Example 7 — Cell F (limiting behaviour of cosec near $0°$ , with picture)

Worked example What happens to

$csc θ$ as $θ$ shrinks toward $0°$ from above? Forecast: we just said $csc 0°$ is undefined. But is the graph calm or wild as we approach it?

Step 1. Sample values: $sin 30° = 0.5 \Rightarrow csc = 2$ ; $sin 10° \approx 0.1736 \Rightarrow csc \approx 5.76$ ; $sin 1° \approx 0.01745 \Rightarrow csc \approx 57.3$ . Why this step? Watching the numbers grow reveals the shape near the forbidden angle.

Step 2. The denominator $sin θ$ gets tinier and tinier, so $1/ sin θ$ gets bigger and bigger without any ceiling. We say $csc θ \to + \infty$ as $θ \to 0^{+}$ . Why this step? "Approaches infinity" describes a trend, not a value at the point. The point itself stays undefined (Example 6).

Reading the figure: the yellow curve is $csc θ$ . The dashed blue lines at $θ = - 180°, 0°, 180°$ are the vertical asymptotes — the forbidden family $θ = k π$ . Notice the curve races up to $+ \infty$ as we approach $0°$ from the right (yellow arrow) and plunges down to $- \infty$ from the left (pink arrow). It never crosses the dashed lines.

Step 3. From below ( $θ \to 0^{-}$ , i.e. a small negative angle), $sin θ < 0$ so $csc θ \to - \infty$ . The graph shoots up on one side and down on the other — a vertical asymptote exactly as defined above.

Verify: $csc 1° = 1/ sin 1° \approx 57.30$ . Compare $csc 0.5° = 1/ sin 0.5° \approx 114.59$ — halving the angle roughly doubled the value, exactly the "blow-up" trend ✓. (This asymptote structure is why $\frac{d}{d θ} csc θ$ in 3.2.04-Derivatives-of-trigonometric-functions also explodes near $0$ .)

Example 8 — Cell F (limiting behaviour of sec and cot at their own forbidden angles)

Worked example Describe the limits of

$sec θ$ as $θ \to 90°$ and of $cot θ$ as $θ \to 180°$ . Forecast: each flip explodes at a different forbidden angle. Predict which value shoots up and which shoots down as you approach from each side.

Step 1 — $sec θ$ near $90°$ . $sec θ = 1/ cos θ$ and $cos 90° = 0$ , so $90° = \frac{π}{2}$ is in $sec$ 's forbidden family. Sample: $cos 89° \approx 0.01745 \Rightarrow sec \approx 57.30$ ; $cos 91° \approx - 0.01745 \Rightarrow sec \approx - 57.30$ . Why this step? Approaching from below $90°$ , $cos$ is small and positive, so $sec \to + \infty$ ; from above, $cos$ is small and negative, so $sec \to - \infty$ . A vertical asymptote at $90°$ .

Step 2 — $cot θ$ near $180°$ . $cot θ = cos θ / sin θ$ and $sin 180° = 0$ , so $180° = π$ is forbidden. Sample: $sin 179° \approx 0.01745$ , $cos 179° \approx - 0.99985 \Rightarrow cot \approx - 57.29$ ; $sin 181° \approx - 0.01745 \Rightarrow cot \approx + 57.29$ . Why this step? Just below $180°$ , $sin$ is small and positive while $cos \approx - 1$ , so $cot \to - \infty$ ; just above, $sin$ flips negative, so $cot \to + \infty$ .

Verify: $sec 89° = 1/ cos 89° \approx 57.2987$ ✓ and $cot 179° = cos 179°/ sin 179° \approx - 57.2900$ ✓ — matching the predicted blow-ups. Every flip has this asymptote picture at its own forbidden family.

Example 9 — Cell G (real-world word problem)

Worked example A wheelchair ramp rises

$1$ m over a horizontal run of $12$ m. A safety code is written in terms of $csc θ$ (hypotenuse per unit rise). Find $csc θ$ and the ramp's actual length. Forecast: which flip involves the hypotenuse and the opposite (rise)? That is $csc$ .

Step 1. Model it as a right triangle: opposite (rise) $a = 1$ m, adjacent (run) $b = 12$ m. Hypotenuse $r = 1^{2} + 1 2^{2} = 145 m \approx 12.042 m .$ Why this step? The Pythagorean theorem gives the slanted ramp length, which the code cares about.

Step 2. $sin θ = \frac{a}{r} = \frac{1}{145}$ , so $csc θ = \frac{r}{a} = \frac{145}{1} = 145 \approx 12.042.$ Why this step? $csc θ = hyp / opp$ is literally "metres of ramp per metre of rise" — exactly the code's quantity.

Verify (units): $csc θ$ is a ratio of two lengths (m/m) so it is dimensionless — good, it equals the ramp length in metres only because the rise happens to be $1$ m. Numerically $145 \approx 12.042$ m, sensibly just longer than the $12$ m run ✓.

Example 10 — Cell H (exam twist: prove an identity)

Consolidate

Recall Which quadrants make each reciprocal negative?

$csc < 0$ where $sin < 0$ (QIII, QIV). $sec < 0$ where $cos < 0$ (QII, QIII). $cot < 0$ where $tan < 0$ (QII, QIV).

Recall Why is

$csc 0°$ "undefined" and not "infinity"? Because $1/0$ has no numerical value; $\infty$ describes the trend as $θ \to 0$ , not the value at $0$ .

Recall State the general forbidden-angle families and each function's period.

$csc, cot$ undefined at $θ = k π$ ; $sec$ undefined at $θ = \frac{π}{2} + k π$ . Periods: $csc$ and $sec$ repeat every $2 π$ ; $cot$ repeats every $π$ .

Every case answered

signs by quadrant (B,C,D), degenerate zeros (E), limiting blow-up for all three flips (F), a physical ramp (G) and an identity proof (H).

Related builds: 2.4.08-Complementary-angle-identities shows why $sec θ = csc (90° - θ)$ ; 2.5.02-Trigonometric-equations-basic uses these flips to solve equations.