Graphs of cosec x, sec x, cot x: Worked examples

This page is a drill through every situation these three graphs can throw at you. We do not learn new theory here — we stress-test the theory from the parent note against every sign, every quadrant, every degenerate case, and a couple of exam-style traps.

The scenario matrix

Every problem in this topic is one (or a combination) of the cells below. Our examples are labelled with the cell they hit, and together they cover the whole grid.

Definition What each cell means (in plain words)

Sign / quadrant — is the input in Q1, Q2, Q3, or Q4? This decides whether $sin, cos$ are $+$ or $-$ , hence the sign of the reciprocal.
Zero / degenerate input — the input lands exactly where the denominator is $0$ . The function is undefined here — you must say so, not guess a number.
Limiting behaviour — the input creeps toward an asymptote ( $x \to nπ$ etc.). We ask which way ( $+ \infty$ or $- \infty$ ).
Touch value — the input lands where $sin / cos = \pm 1$ , so the reciprocal equals $\pm 1$ exactly. (For $cot$ , "touch $\pm 1$ " means the ordinary passing values $cot = \pm 1$ , which are not forbidden-band boundaries — $cot$ has full range.)
Equation / graph-reading — "for what $x$ is $sec x = 2$ ?" — read intersections off the picture.
Word / real-world — a physical setup that hides a reciprocal-trig value.
Exam twist — a trap: forbidden band, wrong asymptote, wrong monotonic direction.

Formula The matrix (rows = case class, cols = which function)

Case class	$csc x$	$sec x$	$cot x$
Q1 value ( $+$ )	Ex 1	Ex 3	Ex 5
Q2/Q3/Q4 sign check	Ex 2	Ex 3	Ex 5
Zero / degenerate (undefined)	Ex 4	Ex 4	Ex 4
Limiting $\to \pm \infty$	Ex 2	Ex 10	Ex 5
Touch / value $\pm 1$	Ex 1	Ex 3	Ex 5
Equation / graph-reading	—	Ex 6	Ex 7
Word / real-world	—	Ex 8	—
Exam twist (forbidden band)	Ex 9	Ex 9	—

Worked examples

Ex 1 — Q1 value and a touch point (cells: Q1 $csc$ , touch $\pm 1$ )

Ex 2 — Q3 sign + limiting direction (cells: Q3 $csc$ , limiting $\to \pm \infty$ )

Ex 3 — $sec$ across quadrants + a touch (cells: $sec$ touch $\pm 1$ , Q2 sign)

Worked example Evaluate

$sec 0$ , $sec \frac{π}{3}$ , and $sec \frac{2 π}{3}$ . Forecast: $cos$ is $+$ in Q1 and $-$ in Q2. So the first two answers are positive, the last negative. One is a touch at $1$ . Guess it.

What: $sec 0 = \frac{1}{cos 0} = \frac{1}{1} = 1$ — a touch (bottom of $sec$ 's upper branch). Note $x = 0$ sits on the boundary between Q4 and Q1 (the positive $x$ -axis), not strictly inside Q1 — it is a boundary/touch case, which is exactly why $cos$ is at its maximum there. Why this step? $cos$ maxes at $1$ at $x = 0$ ; reciprocal is $1$ .
What: $sec \frac{π}{3} = \frac{1}{cos ( π /3 )} = \frac{1}{1/2} = 2$ (this one is genuinely Q1). Why this step? Q1, $cos = \frac{1}{2}$ , positive reciprocal $\geq 1$ .
What: $sec \frac{2 π}{3} = \frac{1}{cos ( 2 π /3 )} = \frac{1}{- 1/2} = - 2$ . Why this step? Q2 cosine is negative, so $sec$ sits on the lower branch.

Verify: $1, 2 \geq 1$ and $- 2 \leq - 1$ ✓; all outside $(- 1, 1)$ ✓.

Ex 4 — The degenerate / undefined cases (cell: zero-input, all three)

Ex 5 — $cot$ sign, touch values, and monotonicity (cells: Q1/Q2 $cot$ , touch $\pm 1$ , limiting)

Worked example Evaluate

$cot \frac{π}{4}$ and $cot \frac{3 π}{4}$ , then state whether $cot x$ is rising or falling on $(0, π)$ and how it behaves as $x \to 0^{+}$ and $x \to π^{-}$ . Forecast: $cot = cos / sin$ . In Q1 both are $+$ (answer $+$ ); in Q2 $cos < 0, sin > 0$ (answer $-$ ). Since $\frac{π}{4} < \frac{3 π}{4}$ and the second answer is smaller, the trend is... guess.

What: $cot \frac{π}{4} = \frac{cos ( π /4 )}{sin ( π /4 )} = \frac{2 /2}{2 /2} = 1$ — an ordinary value of $1$ (for $cot$ , $\pm 1$ are just passing values, not range boundaries: $cot$ has full range). Why this step? At $4 5^{\circ}$ the two components are equal, so their ratio is $1$ .
What: $cot \frac{3 π}{4} = \frac{- 2 /2}{2 /2} = - 1$ — the value $- 1$ , again an ordinary passing value. Why this step? Q2 cosine flips sign; sine stays $+$ .
What (limiting): As $x \to 0^{+}$ , $cos x \to 1$ and $sin x \to 0^{+}$ , so $cot x = \frac{1}{0 ^{+}} \to + \infty$ . As $x \to π^{-}$ , $cos x \to - 1$ and $sin x \to 0^{+}$ , so $cot x = \frac{- 1}{0 ^{+}} \to - \infty$ . Why this step? The reciprocal-of-small-denominator rule again — this is the Vertical Asymptotes and Limits logic applied to $cot$ .
What (monotonicity): Running left to right, $cot$ goes $+ \infty \to 1 \to 0 \to - 1 \to - \infty$ : it is strictly decreasing on $(0, π)$ . Why this step? (standalone WHY): Write $cot x = \frac{cos x}{sin x}$ . On $(0, π)$ the denominator $sin x > 0$ stays positive, while the numerator $cos x$ falls steadily from $1$ down to $- 1$ ; a shrinking top over a top-then-shrinking bottom means the whole ratio only ever drops. Equivalently, its slope (derivative) $- csc^{2} x$ is negative everywhere, so it can never turn back up. That is why $cot$ falls, whereas $tan$ rises.

Verify: $1 > - 1$ while $\frac{π}{4} < \frac{3 π}{4}$ → output fell as input rose → decreasing ✓. Limits: near $0^{+}$ we get $+ \infty$ , near $π^{-}$ we get $- \infty$ ✓, consistent with a curve sliding downhill between two walls.

Ex 6 — Equation, read off the graph (cell: equation-solving $sec$ )

Ex 7 — Where does $cot$ hit zero? (cell: graph-reading $cot$ )

Ex 8 — Real-world word problem (cell: word / $sec$ )

Worked example A ramp rises at angle

$θ$ to the ground. Its slant length $L$ relates to horizontal run $r$ by $L = r sec θ$ . If $r = 3 m$ and $θ = \frac{π}{3}$ , find $L$ . Forecast: $sec θ \geq 1$ always (forbidden band!), so the slant is longer than the run — physically obvious for a slope. Expect $L > 3$ .

What: In the right triangle, $cos θ = \frac{run}{slant} = \frac{r}{L}$ , so $L = \frac{r}{cos θ} = r sec θ$ . Why this step? " $cos =$ adjacent over hypotenuse"; the hypotenuse is the slant. This is exactly why $sec$ shows up in geometry — it answers "how much longer is the hypotenuse than the adjacent side?".
What: $L = 3 \cdot sec \frac{π}{3} = 3 \cdot \frac{1}{cos ( π /3 )} = 3 \cdot \frac{1}{1/2} = 6 m$ . Why this step? Plug $cos \frac{π}{3} = \frac{1}{2}$ .

Verify (units + sanity): metres × dimensionless $sec$ = metres ✓. $6 > 3$ ✓ (slant beats run). And $sec \geq 1$ guarantees $L \geq r$ for any angle — the Domain and Range forbidden band has a physical meaning here.

Ex 9 — Exam twist: the forbidden band (cell: exam twist, $csc / sec$ )

Ex 10 — Limiting behaviour of $sec$ near a wall (cell: limiting $\to \pm \infty$ for $sec$ )

Worked example As

$x \to \frac{π}{2}^{-}$ and $x \to \frac{π}{2}^{+}$ , where does $sec x$ go? Forecast: $x = \frac{π}{2}$ is a $sec$ wall (since $cos = 0$ ). Just left of it $cos > 0$ (Q1), just right $cos < 0$ (Q2). So $sec$ should fly to $+ \infty$ on one side and $- \infty$ on the other. Guess which.

What: As $x \to \frac{π}{2}^{-}$ , $cos x \to 0^{+}$ (small positive, Q1 side), so $sec x = \frac{1}{0 ^{+}} \to + \infty$ . Why this step? Reciprocal of a tiny positive is a huge positive — the same $u \to 0^{+} \Rightarrow 1/ u \to + \infty$ rule from Vertical Asymptotes and Limits.
What: As $x \to \frac{π}{2}^{+}$ , $cos x \to 0^{-}$ (small negative, Q2 side), so $sec x = \frac{1}{0 ^{-}} \to - \infty$ . Why this step? Reciprocal of a tiny negative is a huge negative.

Verify: Sample points: $cos (1.5) \approx 0.0707 > 0$ so $sec (1.5) \approx 14.1 > 0$ (approaching from the left) ✓; $cos (1.6) \approx - 0.0292 < 0$ so $sec (1.6) \approx - 34.2 < 0$ (from the right) ✓. The branch flips sign across the wall exactly like $csc$ did in Ex 2.

Recall Quick self-check across all cells

Sign of $csc$ in Q3? ::: Negative (since $sin < 0$ ), value $\leq - 1$ . Value of $sec 0$ ? ::: $1$ (a touch/boundary point). Is $cot 0$ a number? ::: No — undefined; $x = 0$ is an asymptote. Does $sec x = 0.4$ have a solution? ::: No — $0.4$ is inside the forbidden band $(- 1, 1)$ . As $x \to π^{+}$ , $csc x \to$ ? ::: $- \infty$ (since $sin \to 0^{-}$ ). As $x \to \frac{π}{2}^{+}$ , $sec x \to$ ? ::: $- \infty$ (since $cos \to 0^{-}$ ). Is $cot$ rising or falling on $(0, π)$ ? ::: Falling, from $+ \infty$ to $- \infty$ . For $cot$ , are $\pm 1$ range boundaries? ::: No — $cot$ reaches every real value; $\pm 1$ are ordinary passing values.

Connections

Hinglish version of the parent
Graphs of sin x, cos x, tan x — the waves we reciprocate.
Vertical Asymptotes and Limits — the $1/0 \to \pm \infty$ direction logic (Ex 2, 4, 5, 10).
Trigonometric Identities — reciprocal step used in Ex 6, 9.
Solving Trigonometric Equations — Ex 6, 7, 9.
Domain and Range — the forbidden band $(- 1, 1)$ (Ex 9).
Odd and Even Functions — quadrant-sign reasoning throughout.