Secant, Cosecant, Cotangent and the Inverse Trigonometric Functions

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Two things this lesson assumes — a recap, since neither has its own lesson yet

First: $\cos^2\theta + \sin^2\theta \equiv 1$ , true for every angle $\theta$ , and — like the identities this lesson introduces — it is not printed in the formula booklet. It has to be recalled, not looked up. Alongside it, three shapes worth having ready: $y = \sin\theta$ and $y = \cos\theta$ both oscillate between $-1$ and $1$ and repeat every $360°$ ; $y = \tan\theta$ has no such ceiling — it runs from $-\infty$ to $+\infty$ on each branch, repeats every $180°$ (not $360°$ ), and is undefined wherever $\cos\theta = 0$ , at $\theta = 90°, 270°, \ldots$ , which is exactly where its graph has a vertical .

Second: a function needs to be one-one before it has an inverse. A one-one function never sends two different inputs to the same output; a many-one function sometimes does. The specification states the distinction directly — a function is understood as "a one-one or many-one mapping from ℝ (or a subset of ℝ) to ℝ" — and the reason it matters here is mechanical: if $f$ sent both $30°$ and $150°$ to the same output, an inverse $f^{-1}$ applied to that output would have no way to choose which of the two to send it back to. An inverse is only well-defined on a piece of the domain where the function is one-one.

One consequence of that, stated in the spec's own guidance and worth having ready before it is needed below: "students should know that if $f^{-1}$ exists, then $f^{-1}f(x) = ff^{-1}(x) = x$ " — applying a function and then its inverse, in either order, returns exactly what was started with. That is the test this lesson uses throughout: whatever restricted domain gets chosen for sin, cos or tan, it has to be a domain on which the function is one-one, because that is the only kind of domain on which this round-trip property can hold.

Three new names for three old ratios — and the one trap in the naming

are defined as reciprocals, nothing more: $\sec\theta = \frac{1}{\cos\theta}$ , $\text{cosec}\,\theta = \frac{1}{\sin\theta}$ (also written $\csc\theta$ ), and $\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$ . The spec states this directly as "their relationships to sine, cosine and tangent" — there is no third definition to learn beyond taking a reciprocal.

The trap is entirely in which reciprocal pairs with which name, and it is real: a genuine WMA13 examiner report records candidates who "struggled writing cosec θ = 1/cos θ" — reaching for cosine because "cosec" starts by sounding closer to "cos" than "sin" does. It is the wrong pairing. The letters that matter are the third and fourth: s-e-C pairs with cosine, cos-E-C pairs with sine. Confirm it by checking they still satisfy the same round-trip idea as above: $\sec\theta \cdot \cos\theta = 1$ and $\text{cosec}\,\theta \cdot \sin\theta = 1$ for every value where the original function is non-zero — a reciprocal pair multiplies to $1$ , by definition, and that check catches a mispairing in one line.

Two consequences follow immediately from the definitions and are worth stating before any equation work: first, $\sec\theta$ and $\text{cosec}\,\theta$ are undefined exactly where $\cos\theta$ or $\sin\theta$ is zero — dividing by zero is not a small error to patch, there is no value to assign — and $\cot\theta$ is undefined wherever $\sin\theta = 0$ for the same reason. Second, since $|\cos\theta| \leq 1$ and $|\sin\theta| \leq 1$ always, their reciprocals satisfy $|\sec\theta| \geq 1$ and $|\text{cosec}\,\theta| \geq 1$ always — neither function can ever take a value strictly between $-1$ and $1$ . That is not a separate fact to memorise; it is what taking a reciprocal of a number no bigger than $1$ in size necessarily does.

Turning one true equation into another — by dividing every term by the same thing

In plain terms

Start with three numbers that satisfy a special relationship: 9, 16 and 25, where 9 + 16 = 25 (they're the squares of 3, 4 and 5 — the sides of a right-angled triangle). Now divide every one of those three numbers by 9: 9 ÷ 9 = 1, 16 ÷ 9 = 1.78 (to 2 d.p.), and 25 ÷ 9 = 2.78 (to 2 d.p.). Check the new equation still holds: 1 + 1.78 = 2.78. ✓ It does — dividing all three numbers in a true equation by the same number (9, here) always produces another true equation, just with different-looking numbers in it. Now try dividing the ORIGINAL three numbers by 16 instead: 9 ÷ 16 = 0.5625, 16 ÷ 16 = 1, 25 ÷ 16 = 1.5625. Check: 0.5625 + 1 = 1.5625. ✓ Also true — a completely different-looking equation, built from the exact same starting fact, just divided through by a different number this time.

Those three starting numbers, 9, 16 and 25, aren't random — they're $\cos^2\theta$ , $\sin^2\theta$ and $1$ for one specific angle θ, scaled up to clear the fractions: a 3-4-5 triangle gives $\cos\theta = \frac{3}{5}$ and $\sin\theta = \frac{4}{5}$ , so $\cos^2\theta = \frac{9}{25}$ , $\sin^2\theta = \frac{16}{25}$ , and indeed $\frac{9}{25} + \frac{16}{25} = 1$ . " $9 + 16 = 25$ " is a scaled-up copy of the identity $\cos^2\theta + \sin^2\theta \equiv 1$ , which is true for every angle, not just this one triangle. Dividing every term by 9 (i.e. by $\cos^2\theta$ ) is the exact same move as dividing every term of the identity by $\cos^2\theta$ — and the two numbers that changed, 1.78 and 2.78, are, in disguise, $\tan^2\theta$ and $\sec^2\theta$ for this angle: $\tan\theta = \frac{4}{3}$ , so $\tan^2\theta = \frac{16}{9} \approx 1.78$ ; $\sec\theta = \frac{5}{3}$ , so $\sec^2\theta = \frac{25}{9} \approx 2.78$ . Dividing by 16 (i.e. by $\sin^2\theta$ ) instead is the same move that produces the companion identity, with $\cot^2\theta$ and $\text{cosec}^2\theta$ standing where $\tan^2\theta$ and $\sec^2\theta$ stood a moment ago.

Formally

Formally: start from $\cos^2\theta + \sin^2\theta \equiv 1$ , true for every $\theta$ , and divide every term through by $\cos^2\theta$ (valid wherever $\cos\theta \neq 0$ ): $\frac{\cos^2\theta}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}$ , which simplifies to $1 + \tan^2\theta \equiv \sec^2\theta$ . Dividing the same starting identity by $\sin^2\theta$ instead gives the companion identity, $\cot^2\theta + 1 \equiv \text{cosec}^2\theta$ — the full general derivation for both, worked through one careful step at a time, follows next.

Mechanism

Where sec²θ = 1 + tan²θ actually comes from

Start from the one identity that is genuinely foundational — $\cos^2\theta + \sin^2\theta \equiv 1$ — and divide every term by $\cos^2\theta$ (valid wherever $\cos\theta \neq 0$ , which is exactly where $\sec\theta$ is defined in the first place, so nothing is lost). $\frac{\cos^2\theta}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}$ . The first term is $1$ . The second term is $\left(\frac{\sin\theta}{\cos\theta}\right)^2 = \tan^2\theta$ . The right-hand side is $\left(\frac{1}{\cos\theta}\right)^2 = \sec^2\theta$ . So the division produces, without any further assumption, $1 + \tan^2\theta \equiv \sec^2\theta$ — the identity was already sitting inside $\cos^2\theta + \sin^2\theta \equiv 1$ , waiting to be uncovered by dividing through by the right thing. Divide the SAME starting identity by $\sin^2\theta$ instead, and the companion identity falls out the same way: $\frac{\cos^2\theta}{\sin^2\theta} + \frac{\sin^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta}$ , giving $\cot^2\theta + 1 \equiv \text{cosec}^2\theta$ . Both identities are the same starting fact, viewed through two different divisions — which is also why they are not independent things to memorise separately so much as one thing to know how to re-derive twice. This matters for exam technique specifically: general marking guidance for this paper records that where a method uses a learned formula or identity, examiners advise quoting it BEFORE substituting values, because the method mark can otherwise only be inferred from correct working and is then vulnerable to any slip inside that working. Writing " $\sec^2\theta \equiv 1 + \tan^2\theta$ " as its own line, before using it, is the safer habit — and it is genuinely the same advice this course's own quadratic-functions lesson gives for quoting the quadratic formula before substituting into it.

Diagram — sec θ, cosec θ and cot θ — asymptotes where the original function's own zeros were

x-axis: θ (degrees, 0° to 360°) · y-axis: y = f(θ)

y = sec θ: Undefined and asymptotic wherever cos θ = 0 — at θ = 90° and θ = 270°. Between the asymptotes it forms U-shaped branches: a branch opening upward with minimum value 1 where cos θ = 1 (θ = 0°, 360°), and a branch opening downward with maximum value −1 where cos θ = −1 (θ = 180°). Never takes a value strictly between −1 and 1.
y = sec θ (cont.): Undefined and asymptotic wherever cos θ = 0 — at θ = 90° and θ = 270°. Between the asymptotes it forms U-shaped branches: a branch opening upward with minimum value 1 where cos θ = 1 (θ = 0°, 360°), and a branch opening downward with maximum value −1 where cos θ = −1 (θ = 180°). Never takes a value strictly between −1 and 1.
y = sec θ (cont. 2): Undefined and asymptotic wherever cos θ = 0 — at θ = 90° and θ = 270°. Between the asymptotes it forms U-shaped branches: a branch opening upward with minimum value 1 where cos θ = 1 (θ = 0°, 360°), and a branch opening downward with maximum value −1 where cos θ = −1 (θ = 180°). Never takes a value strictly between −1 and 1.
y = cosec θ: Undefined and asymptotic wherever sin θ = 0 — at θ = 0°, 180° and 360°. Same U-shaped branch structure as sec θ, shifted: a downward-opening branch (maximum −1) between 0° and 180° where sin θ is negative-to-positive... more precisely, an upward branch with minimum 1 centred near θ = 90° (where sin θ = 1) and a downward branch with maximum −1 centred near θ = 270° (where sin θ = −1). Never strictly between −1 and 1.
y = cosec θ (cont.): Undefined and asymptotic wherever sin θ = 0 — at θ = 0°, 180° and 360°. Same U-shaped branch structure as sec θ, shifted: a downward-opening branch (maximum −1) between 0° and 180° where sin θ is negative-to-positive... more precisely, an upward branch with minimum 1 centred near θ = 90° (where sin θ = 1) and a downward branch with maximum −1 centred near θ = 270° (where sin θ = −1). Never strictly between −1 and 1.
y = cot θ: Undefined and asymptotic wherever sin θ = 0 — the same locations as cosec θ: 0°, 180°, 360°. Unlike sec θ and cosec θ, cot θ is not bounded away from the strip between −1 and 1 — like tan θ, whose reciprocal it is, it takes every real value on each branch, decreasing continuously from +∞ down to −∞ between consecutive asymptotes.
y = cot θ (cont.): Undefined and asymptotic wherever sin θ = 0 — the same locations as cosec θ: 0°, 180°, 360°. Unlike sec θ and cosec θ, cot θ is not bounded away from the strip between −1 and 1 — like tan θ, whose reciprocal it is, it takes every real value on each branch, decreasing continuously from +∞ down to −∞ between consecutive asymptotes.

Asymptote rule: A reciprocal function's asymptotes sit exactly where the original function's own graph crosses zero — because dividing by a number approaching zero sends the reciprocal to infinity. sec θ's asymptotes come from cos θ = 0; cosec θ's and cot θ's come from sin θ = 0. The function being reciprocated decides the asymptote locations, not the reciprocal function's own name.
Range of sec θ and cosec θ: y ≤ −1 or y ≥ 1, for both — direct consequence of |cos θ| ≤ 1 and |sin θ| ≤ 1 meaning their reciprocals have magnitude at least 1.
Range of cot θ: All real y — unbounded, like tan θ. Reciprocating tan θ does not bound it the way reciprocating cos θ or sin θ does, because tan θ itself is already unbounded near its own asymptotes, so 1/tan θ passes smoothly through 0 rather than being pushed away from a bounded strip.
Period: sec θ and cosec θ repeat every 360°, matching cos θ and sin θ. cot θ repeats every 180°, matching tan θ — not 360°.

Common error: Drawing y = sec θ with asymptotes at 0°, 180°, 360° — the locations that actually belong to cosec θ (and cot θ).

Correct: sec θ's asymptotes come from cos θ = 0 (90°, 270°); cosec θ's and cot θ's come from sin θ = 0 (0°, 180°, 360°). This is the graphical form of the same sec/cosec naming confusion documented directly on a WMA13 paper: candidates who write cosec θ = 1/cos θ would, if asked to sketch it, draw the asymptotes in the wrong place for exactly the same underlying reason.

Mechanism

Why arcsin, arccos and arctan need restricted domains at all

None of sin, cos or tan is over its full domain — each repeats every period, so each is , and a many-one function has no inverse until its domain is cut down to a one-one piece first. The three restricted domains used for are not arbitrary; each is chosen as the specific interval containing θ = 0 (or as close to it as the function allows) on which the function is monotonic — strictly increasing or strictly decreasing, hence one-one — and covers the function's full range while doing so. For sin θ, that interval is $-90° \leq \theta \leq 90°$ : sin θ increases continuously from $-1$ to $1$ across it, touching every output value in $[-1, 1]$ exactly once, so $\arcsin x$ is defined with domain $[-1, 1]$ and range $-90° \leq \arcsin x \leq 90°$ — the restricted domain of sin θ becomes the range of its inverse, and the restricted range of sin θ (also $[-1,1]$ , since that IS the range of sin θ everywhere) becomes the domain of arcsin, by the general rule that a function's domain and range swap places under inversion. For cos θ, the corresponding interval is $0° \leq \theta \leq 180°$ : cos θ decreases continuously from $1$ to $-1$ , one-one across the whole interval, so $\arccos x$ has domain $[-1, 1]$ and range $0° \leq \arccos x \leq 180°$ . For tan θ the story has one extra piece: tan θ is undefined at $\pm 90°$ itself, so the interval is the OPEN interval $-90° < \theta < 90°$ , across which tan θ increases from $-\infty$ to $+\infty$ , one-one and covering every real output — so $\arctan x$ has domain all of $\mathbb{R}$ (every real number is achieved) and range the open interval $-90° < \arctan x < 90°$ , never reaching either endpoint, which is why $y = \arctan x$ has two horizontal asymptotes rather than two endpoints. A domain question about $\arcsin(1.5)$ is therefore not asking for a value at all: $1.5$ is outside $[-1, 1]$ , the range sin θ can ever produce, so there is no input to invert and the expression is undefined — exactly the same kind of domain check the spec expects for any inverse function, applied here to a specific trio of restricted domains rather than left as an abstract rule.

Diagram — arcsin, arccos and arctan — the restricted domains printed straight onto the inverse graph

x-axis: x · y-axis: y = f⁻¹(x), in degrees

y = arcsin x: Domain −1 ≤ x ≤ 1, range −90° ≤ y ≤ 90°. Strictly increasing, passing through (0, 0°), (1, 90°) and (−1, −90°) — the mirror image of y = sin θ on −90° ≤ θ ≤ 90°, reflected in the line y = x, which is what taking an inverse does to any one-one function's graph.
y = arccos x: Domain −1 ≤ x ≤ 1, range 0° ≤ y ≤ 180°. Strictly decreasing, passing through (0, 90°), (1, 0°) and (−1, 180°) — the reflection of y = cos θ on 0° ≤ θ ≤ 180° in the line y = x.
y = arctan x: Domain all real x, range −90° < y < 90° (open, never reached). Strictly increasing, passing through (0, 0°); horizontal asymptotes at y = 90° and y = −90° as x → +∞ and x → −∞, the reflection of tan θ's own vertical asymptotes at θ = ±90° on the restricted branch.

Domain of arcsin/arccos: [−1, 1]: Because sin θ and cos θ themselves never leave [−1, 1] — there is no θ for which sin θ = 1.5, so arcsin(1.5) has nothing to invert and is undefined, not merely large.
Domain of arctan: all of ℝ: Because tan θ, on its restricted branch, already reaches every real number — there is no output of tan θ left over for arctan to be unable to invert.
Reflection in y = x: Every inverse function's graph is its original's graph reflected in the line y = x — swapping the roles of input and output is exactly what 'inverse' means, and the restricted domain chosen for sin/cos/tan is what determines the shape of that reflection here.
arctan's asymptotes, not endpoints: y = arctan x approaches but never reaches 90° or −90°, because tan θ itself is undefined at those two angles — there is no input that maps to exactly ±90°, so arctan can never output it.

Common error: Sketching y = arcsin x as if it continued past x = 1 or x = −1, or evaluating arcsin(1.5) as some large angle.

Correct: arcsin is undefined outside [−1, 1] — full stop, not a large or approximate value. This is a domain restriction on the INPUT, not a range restriction on the output, and it is the identical failure mode documented directly on real inverse-function questions on this paper: candidates who find a correct expression for an inverse but do not check or state the domain it is actually valid on.

In your own words

In one sentence: why can $\sec\theta$ and $\text{cosec}\,\theta$ never take a value strictly between $-1$ and $1$ , while $\cot\theta$ can take any real value at all?

Worked, in full

The full solution set for sec²θ = 3 + tanθ, 0° ≤ θ < 360° — including the branch it's easy to stop after finding the nice one

01
Rewrite sec²θ using the identity derived above, so the equation is in terms of a single function. $\sec^2\theta \equiv 1 + \tan^2\theta$ , so $1 + \tan^2\theta = 3 + \tan\theta$ , and collecting everything on one side: $\tan^2\theta - \tan\theta - 2 = 0$ .
Earns: M1 — uses sec²θ ≡ 1 + tan²θ to rewrite the equation as a 3-term quadratic in tan θ alone. This is the step the equation is actually testing: recognising that sec²θ is not a separate unknown but a disguised form of tan²θ.
02
Solve the quadratic in tan θ. $\tan^2\theta - \tan\theta - 2 = 0$ factorises as $(\tan\theta - 2)(\tan\theta + 1) = 0$ , giving $\tan\theta = 2$ or $\tan\theta = -1$ .
Earns: dM1 — solves the resulting 3-term quadratic by any of the credited routes (factorisation, formula, or completing the square), dependent on the M1 above. This is the same general marking principle the quadratic-functions lesson uses: whichever route reaches the correct pair of values earns this mark.
03
tan θ = −1 is a standard angle: the reference angle is 45°, and tan θ is negative in the second and fourth quadrants, giving θ = 180° − 45° = 135° and θ = 360° − 45° = 315°.
Earns: A1 — both exact solutions from the tan θ = −1 branch, 135° and 315°.
04
tan θ = 2 is NOT a standard angle: the reference angle is $\arctan 2 = 63.4°$ (1 d.p.), and tan θ is positive in the first and third quadrants, giving θ = 63.4° and θ = 180° + 63.4° = 243.4°.
Earns: A1 — both solutions from the tan θ = 2 branch, correct to 1 decimal place, in both quadrants where tangent is positive.
05
State the complete solution set: θ = 63.4°, 135°, 243.4°, 315°. All four belong to the same equation; stopping after the exact-looking branch (135°, 315°) and never returning to the branch that needed a calculator would silently discard exactly half of a fully correct method.
Earns: Nothing further — the mark scheme has run out by this stage. It earns nothing and costs nothing to write, and it is the line that catches a dropped branch before the answer is submitted.

Source — Examiner report, Jun 2022

"It was disappointing that many candidates failed to identify the second solution here to gain the full marks"

Complete it yourself

Complete the chain — the full solution set for cosec²θ + cotθ = 3, 0° ≤ θ < 360°

01
Rewrite cosec²θ using cosec²θ ≡ 1 + cot²θ, the companion identity derived the same way as sec²θ ≡ 1 + tan²θ (dividing cos²θ + sin²θ ≡ 1 by sin²θ instead of cos²θ). Substituting: $1 + \cot^2\theta + \cot\theta = 3$ .
02
Collect onto one side to get a 3-term quadratic in cot θ: $\cot^2\theta + \cot\theta - 2 = 0$ . Here a = 1, b = 1, c = −2.

Same question, every valid method

Solve, for 0° ≤ θ < 360°, the equation cosec²θ − cotθ = 7, giving non-exact answers to 1 decimal place. (VERIDIAN-original question — not a reproduction of any past-paper question. The general marking principle demonstrated here, that a 3-term quadratic can be solved by any of several credited routes for the same marks, is verified from real WMA13 general marking guidance.)

2 valid methods · every one reaches θ = 18.4°, 153.4°, 198.4°, 333.4° · 4 marks available

01
$\text{cosec}^2\theta \equiv 1 + \cot^2\theta \Rightarrow 1 + \cot^2\theta - \cot\theta = 7 \Rightarrow \cot^2\theta - \cot\theta - 6 = 0$
Method mark for using the identity to form a 3-term quadratic in cot θ alone. This is the step the question is actually testing — everything after this line is routine quadratic-solving.
M1
02
$(\cot\theta - 3)(\cot\theta + 2) = 0 \Rightarrow \cot\theta = 3 \text{ or } \cot\theta = -2$
Method mark for solving the quadratic by factorisation — the two factors multiply to give −6 (the constant term) and add to give −1 (the coefficient of cot θ), confirming the factor pair is correct before it is used.
M1
03
$\cot\theta = 3 \Rightarrow \tan\theta = \frac{1}{3} \Rightarrow \theta = 18.4°$ or $\theta = 180° + 18.4° = 198.4°$
Accuracy mark for both solutions from the cot θ = 3 branch, correct to 1 d.p. Tan θ is positive here, so both solutions fall in the quadrants where tangent is positive.
A1
04
$\cot\theta = -2 \Rightarrow \tan\theta = -0.5 \Rightarrow \theta = 180° - 26.6° = 153.4°$ or $\theta = 360° - 26.6° = 333.4°$
Accuracy mark for both solutions from the cot θ = −2 branch. Missing either branch, or only one solution within a branch, is the single most common way to lose a mark on this question type.
A1

Fastest whenever the quadratic's factor pair is visible by inspection, as it is here: −6 splits as −3 × 2, and −3 + 2 = −1 matches the middle coefficient immediately. Its one risk is the same as for any quadratic: if no integer factor pair exists, searching for one has no natural stopping point, and the quadratic formula below should be reached for instead.

Marked, line by line

θ is obtuse, and sinθ = 4/5. (a) Find the exact value of secθ and the exact value of cotθ. (3) (b) Hence find, for 0° ≤ x < 360°, all values of x for which cot x = cotθ, giving your answers to 1 decimal place. (4) — VERIDIAN-original question, inspired by the structure of real WMA13 items that give a ratio plus a quadrant and ask for other exact ratios, then extend into an equation with a "hence" part (spec 2.1 and 2.2 in one question). Not a reproduction of any past-paper question; several genuine WMA13 sub-parts carry the instruction "In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable," verbatim from the January 2023 question paper, attached to parts exactly like part (a) where a calculator would trivially bypass the algebra being examined.

7 marks available

(a) — 3 marks

01
$\cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$ , so $\cos\theta = \pm\frac{3}{5}$
Method mark for using cos²θ + sin²θ ≡ 1 to attempt cos θ. Earned for the attempt at this identity, regardless of which sign is chosen next.
M1
02
θ is obtuse ( $90° < \theta < 180°$ ), where cosine is negative, so $\cos\theta = -\frac{3}{5}$ , giving $\sec\theta = \frac{1}{\cos\theta} = -\frac{5}{3}$
Accuracy mark, correct answer only. This is the mark that tests whether the given quadrant was actually used: the algebra of the identity alone cannot decide the sign of cos θ, only the quadrant can, and the quadrant was stated in the question for exactly this reason.
A1
03
$\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{-3/5}{4/5} = -\frac{3}{4}$
Accuracy mark, follow through their cos θ. Whatever sign was chosen for cos θ on the line above, this mark rewards correctly dividing it by the given sin θ = 4/5 — the method (cos θ over sin θ) is what is being tested here, applied to their own value.
A1 ft

(b) — 4 marks

101
$\cot x = \cot\theta \Leftrightarrow \tan x = \frac{1}{\cot\theta}$ , so using their cot θ from (a): $\tan x = \frac{1}{-3/4} = -\frac{4}{3}$
Method mark for rewriting the condition on cot x as an equivalent condition on tan x, applied to whichever cot θ value part (a) produced. Plain M1, not 'M1 ft' — an M mark rewards the method itself, not the value it's applied to, so it needs no follow-through tag to survive an earlier slip; only A and B marks in real WMA13 mark schemes are ever tagged ft.
M1
102
Reference angle: $\arctan\left(\frac{4}{3}\right) = 53.1°$ (1 d.p.)
Method mark for finding the reference angle from the magnitude of tan x, independent of its sign — the sign is applied in the next two lines.
M1
103
$\tan x$ is negative, so $x$ lies in the second or fourth quadrant: $x = 180° - 53.1° = 126.9°$
Accuracy mark, follow through their sign of tan x, for the second-quadrant solution.
A1 ft
104
$x = 360° - 53.1° = 306.9°$
Accuracy mark, follow through, for the fourth-quadrant solution. Both quadrants are required — stopping after one is the same error as stopping after one branch of a quadratic.
A1 ft

Named traps

cosec-written-as-reciprocal-of-cosine: Confirmed directly on a real WMA13 identity question: candidates "struggled writing cosec θ = 1/cos θ" (Jan 2022, Q2) — reaching for cosine because of the shared first syllable, when the correct pairing is cosec θ = 1/sin θ. The same report records the flip side too: "even some of the weakest candidates were able to earn a mark for stating the identity cosec θ = 1/sin θ" (Jan 2022, Q2) — this is a genuinely accessible mark, and losing it to a mispairing rather than a harder step is the most avoidable loss in this whole topic. Check any pairing by confirming it multiplies to 1: cosec θ · sin θ = 1, not cosec θ · cos θ.
coefficient-folded-into-the-reciprocal: A second, distinct error on the exact same real WMA13 question named above, re-verified directly against the primary mark scheme (not just the examiner-report prose): the same series' report records candidates "sometimes" writing "3 cosecθ = 1/ 3sinθ" (Jan 2022, Q2) — folding the coefficient into the denominator alongside sin θ, rather than leaving it multiplying the finished reciprocal. The real mark scheme is explicit that this scores nothing for the identity mark: "Note that 3cosecθ = 1/(3sinθ) is B0 unless there is an aside that does state cosecθ = 1/sinθ." The fix is mechanical, not a new idea: reciprocate the trig function alone first — cosec θ = 1/sin θ — then multiply by whatever coefficient sits in front, e.g. 3 cosec θ = 3 × (1/sin θ) = 3/sin θ. The coefficient never moves inside the denominator next to sin θ.
quadratic-in-tan-or-cot-missing-a-second-solution: Documented on a different WMA13 trig equation, not one reducing via sec²θ or cosec²θ specifically, but the exact same shape of failure: "it was disappointing that many candidates failed to identify the second solution here to gain the full marks" (Jun 2022, Q7). Every equation in this lesson that reduces to a quadratic in tan θ or cot θ produces TWO values of the trig ratio, and each of those typically produces TWO angles in a full 0°–360° range — up to four solutions from one equation. Stopping after the first branch, or after one solution within a branch, is this trap in miniature.
solved-ratio-mistaken-for-the-angle-itself: Confirmed on a WMA13 question that solved for sin x rather than x directly: "a few interpreted the solution to their equation as being the value for x rather than for sin x, thereby losing both marks" (Jan 2024, Q6(c)). The identical failure is available here with tan θ or cot θ: solving cot²θ − cotθ − 6 = 0 gives values of cot θ (3 and −2), not values of θ. The step from "cot θ = 3" to "θ = 18.4° or 198.4°" is not optional bookkeeping — it is a required, markable step that a genuine number of candidates skip on the analogous sin-based question.
identity-not-quoted-before-use: General marking guidance for this paper states the advice explicitly: "where a method involves using a formula that has been learnt, the advice given in recent examiners' reports is that the formula should be quoted first... where the formula is not quoted, the method mark can be gained by implication from correct working with values but may be lost if there is any mistake in the working" (Jan 2023 general marking guidance, cross-checked against Oct 2023 and Jun 2022). Applied here: writing "sec²θ ≡ 1 + tan²θ" as its own line before substituting is what makes the method mark secure even if the next line contains a slip — skipping straight to the substituted equation makes the method mark depend on everything downstream staying correct.
restricted-domain-of-the-inverse-left-unstated: Confirmed as the single most consistently dropped mark on real inverse-function questions on this paper: "despite being a standard question, the majority of candidates still failed to state the domain for their inverse function and did not achieve the B mark" (Jun 2022, Q2), and independently, "a majority of candidates were not aware that the domain was required, and therefore by far the most common score seen was 2/3" (Jan 2022, Q6). Those quotes are about inverse functions generally (spec 1.2), not about arcsin/arccos/arctan specifically — but the failure is identical in shape: a question asking for the domain OR range of arcsin, arccos or arctan is asking for exactly the kind of stated restriction these reports record candidates omitting, even when the rest of the working is correct.
missing-intermediate-line-in-a-prove-that: A general pattern confirmed across a WMA13 series' full set of reports: "it was noticeable in this series that many candidates omitted important lines when proceeding to the given solution resulting in the loss of some vital marks" (Jan 2022, general summary). Applied to this topic specifically: a "show that sec²θ ≡ 1 + tan²θ" question is asking for the division-by-cos²θ derivation shown above written out, not just asserted — jumping from cos²θ + sin²θ ≡ 1 straight to the answer, with the dividing-through step invisible, is exactly the omission these reports repeatedly penalise.

Seven traps, one habit each — the checklist before the graded questions

Each trap above has a single concrete habit that closes it, and none of the seven require new mathematics beyond what's already been worked through — only doing the step that's easiest to skip once the harder part is finished. Before attempting the MCQs below, run through all seven against whatever reciprocal or inverse-trig question is in front of you.

About to write cosec θ as a reciprocal? Say the pairing out loud before writing the fraction: s-e-C pairs with cos, cos-E-C pairs with sin. This is the one mark real candidates lose to a first-syllable guess rather than to any actual mathematics.

Is there a coefficient in front — 3 cosec θ, not just cosec θ? Reciprocate the trig function alone first, then multiply by the coefficient: 3 cosec θ = 3 × (1/sin θ) = 3/sin θ. A real WMA13 mark scheme scores 3 cosecθ = 1/(3 sinθ) as earning nothing, because the coefficient has been folded into the denominator instead of left outside it.

Reached a quadratic in tan θ or cot θ? Expect two values of the ratio, and expect each one to produce two angles in a full 0°–360° sweep — up to four solutions from one equation. Write out both branches before declaring the equation solved, even when the first branch alone looks like a complete answer.

Just solved for tan θ or cot θ, not θ itself? That value is not the answer — it's the input to one more step. Write the line that turns "cot θ = 3" into "θ = 18.4° or 198.4°" explicitly; it is a required, markable step, not bookkeeping to skip.

About to use sec²θ ≡ 1 + tan²θ or cosec²θ ≡ 1 + cot²θ? Quote the identity on its own line before substituting into it. Neither is in the formula booklet, and quoting first is what keeps the method mark secure even if a later line slips.

Asked for the domain or range of arcsin, arccos or arctan? State it explicitly, as its own line — real candidates on this paper lose this mark more consistently than any other on inverse-function questions, even when every other line of working is correct.

Asked to show that an identity holds? Write out the actual division-by-cos²θ-or-sin²θ step, not just the identity it produces. A "show that" question is asking for the intermediate line to be visible, not merely asserted as the final answer.

Retrieval — with feedback on every choice

Question 1

2 marks

Given that $\cos\theta = -\frac{2}{5}$ , find the exact value of $\sec\theta$ .

Question 2

2 marks

For $0° \leq \theta < 360°$ , at which values does $y = \text{cosec}\,\theta$ have vertical asymptotes?

Question 3

2 marks

What is $\arcsin(1.5)$ ?

Question 4

1 mark

Which of these correctly defines $\cot\theta$ ?

Question 5

3 marks

How many solutions does $\text{cosec}^2\theta = 2$ have for $0° \leq \theta < 360°$ ?

Question 6

1 mark

What is the range of $y = \arctan x$ ?

Question 7

1 mark

Which of these correctly rewrites $4\,\text{cosec}\,\theta$ as a single fraction in terms of $\sin\theta$ ?

Reference — not a study method, a lookup

sec θ = 1/cos θ; cosec θ = 1/sin θ; cot θ = cos θ/sin θ = 1/tan θ. Not the other pairing.
A coefficient in front stays outside the reciprocal: 3 cosec θ = 3/sin θ, never 1/(3 sin θ) — a real, mark-scheme-penalised WMA13 slip.
sec²θ ≡ 1 + tan²θ; cosec²θ ≡ 1 + cot²θ — both from dividing cos²θ+sin²θ≡1 by cos²θ or sin²θ. Memorise; not in the booklet.
sec θ, cosec θ: |value| ≥ 1 always; undefined where cos θ, sin θ = 0. cot θ: any real value; undefined where sin θ = 0.
arcsin: domain [−1,1], range −90°–90°. arccos: domain [−1,1], range 0°–180°. arctan: domain ℝ, range −90°<y<90°, open.
Inverse exists only where the original is one-one — that's why each range above is a restriction, not the full curve.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. The mark allocations attached to this lesson's MCQs, marked-solution and method-comparison are VERIDIAN-original, modelled on real mark-scheme conventions rather than transcribed from an actual Pearson mark scheme — there is no real mark scheme for a question that isn't a real past-paper question.

Question 12 marks

Given that $\cos\theta = -\frac{2}{5}$ , find the exact value of $\sec\theta$ .

$-\frac{5}{2}$
Correct. $\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-2/5} = -\frac{5}{2}$ — reciprocate the whole fraction, sign included.
B $\frac{5}{2}$
The magnitude is right and the sign has been dropped. $\sec\theta$ takes the same sign as $\cos\theta$ always, since dividing 1 (positive) by a negative number gives a negative result.
C $-\frac{2}{5}$
This restates $\cos\theta$ without taking the reciprocal at all. $\sec\theta$ is a genuinely different value from $\cos\theta$ , not another name for the same number.
D $\frac{2}{5}$
Both the sign and the reciprocal step are missing — this is $|\cos\theta|$ , neither reciprocated nor correctly signed.

Traps tested: Sign dropped on reciprocal · Reciprocal step skipped

Question 22 marks

For $0° \leq \theta < 360°$ , at which values does $y = \text{cosec}\,\theta$ have vertical asymptotes?

$\theta = 0°, 180°$ (and the curve is undefined again as $\theta \to 360°$ )
Correct. $\text{cosec}\,\theta = \frac{1}{\sin\theta}$ is undefined wherever $\sin\theta = 0$ , which happens at $\theta = 0°, 180°, 360°$ — the zeros of $\sin\theta$ itself, not of $\cos\theta$ .
B $\theta = 90°, 270°$
These are the zeros of $\cos\theta$ , and therefore the asymptotes of $\sec\theta$ , not $\text{cosec}\,\theta$ . This is the exact sec/cosec pairing confusion documented on this paper — a reciprocal's asymptotes come from the ORIGINAL function's zeros, and the original function here is $\sin\theta$ , not $\cos\theta$ .
C $\theta = 45°, 135°, 225°, 315°$
These are not zeros of $\sin\theta$ at all — $\sin\theta$ is $\pm\frac{\sqrt{2}}{2}$ at each of these, nowhere near zero, so $\text{cosec}\,\theta$ is perfectly well-defined there.
D $\text{cosec}\,\theta$ has no asymptotes — it is defined for every value of $\theta$
$\text{cosec}\,\theta$ is a reciprocal, and a reciprocal is undefined wherever the original function is zero. $\sin\theta$ is zero three times in this range, so $\text{cosec}\,\theta$ is undefined (and asymptotic) at each of them.

Traps tested: Cosec asymptotes confused with secant · Asymptote locations invented · Reciprocal assumed always defined

Question 32 marks

What is $\arcsin(1.5)$ ?

Undefined — $1.5$ is outside the domain $[-1, 1]$ of $\arcsin$
Correct. $\sin\theta$ never exceeds $1$ or goes below $-1$ for any real $\theta$ , so there is no angle whose sine is $1.5$ — the domain restriction on $\arcsin$ is not a technicality, it is a direct statement of what $\sin\theta$ can and cannot produce.
B $90°$
$90°$ is $\arcsin(1)$ , the boundary of the domain, not $\arcsin(1.5)$ . Confusing the input with the boundary value nearest to it treats the domain restriction as soft when it is exact.
CApproximately $48.6°$
This looks like a calculator answer, and a genuine calculator forced to evaluate this will typically return a math-domain error rather than a number — there is no real angle to find, so no numerical answer, however plausible-looking, can be correct here.
DThere is no such function as $\arcsin$ , only $\sin$
$\arcsin$ is a genuine, well-defined function, with domain $[-1, 1]$ and range $-90°$ to $90°$ — the issue here is not that $\arcsin$ doesn't exist, it is that $1.5$ is not in the interval where it is defined.

Traps tested: Domain boundary confused with input · Domain restriction ignored · Arcsin existence denied

Question 41 mark

Which of these correctly defines $\cot\theta$ ?

$\frac{\cos\theta}{\sin\theta}$
Correct — equivalently $\frac{1}{\tan\theta}$ , since $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\cot\theta$ is its reciprocal.
B $\frac{\sin\theta}{\cos\theta}$
This is $\tan\theta$ , not $\cot\theta$ — the fraction has been inverted. Cotangent is the reciprocal of tangent, so it needs cosine on top, sine on the bottom, the opposite way round from tangent itself.
C $\frac{1}{\cos\theta}$
This is $\sec\theta$ , the reciprocal of cosine, not the reciprocal of tangent.
D $\frac{1}{\sin\theta}$
This is $\text{cosec}\,\theta$ , the reciprocal of sine, not the reciprocal of tangent.

Traps tested: Cot defined as tan · Cot defined as sec · Cot defined as cosec

Question 53 marks

How many solutions does $\text{cosec}^2\theta = 2$ have for $0° \leq \theta < 360°$ ?

$4$
Correct. $\text{cosec}^2\theta = 2 \Rightarrow \sin^2\theta = \frac{1}{2} \Rightarrow \sin\theta = \pm\frac{\sqrt{2}}{2}$ — TWO values of $\sin\theta$ , from the squared equation, each giving two solutions in range: $\sin\theta = \frac{\sqrt{2}}{2}$ gives $\theta = 45°, 135°$ ; $\sin\theta = -\frac{\sqrt{2}}{2}$ gives $\theta = 225°, 315°$ . Four solutions in total.
B $2$
This counts only the $\sin\theta = +\frac{\sqrt{2}}{2}$ branch and drops the negative branch entirely. Squaring $\sin\theta$ to get $\sin^2\theta = \frac{1}{2}$ loses the sign of $\sin\theta$ , so BOTH square roots — positive and negative — are genuine solutions and both need to be checked.
C $1$
This treats the equation as though it had one solution the way a linear equation would. A squared trig equation of this shape has two values of the ratio and typically four angles in a full period — dropping to a single answer discards nearly all of them.
D $8$
This double-counts: each of the four genuine solutions has been counted twice, perhaps by treating $+\frac{\sqrt{2}}{2}$ and $-\frac{\sqrt{2}}{2}$ as each producing four angles rather than two.

Traps tested: Negative root branch dropped · Squared equation treated as single valued · Solutions double counted

Question 61 mark

What is the range of $y = \arctan x$ ?

$-90° < y < 90°$ , approached but never reached
Correct. $\tan\theta$ itself is undefined at $\theta = \pm 90°$ , so there is no input to $\arctan$ that maps to exactly $90°$ or $-90°$ — the curve has horizontal asymptotes there instead of endpoints.
B $-90° \leq y \leq 90°$ , including both endpoints
The endpoints are exactly what is excluded. Including them would require some real $x$ with $\arctan x = 90°$ , which would mean $\tan 90°$ is defined and equals that $x$ — but $\tan\theta$ has no value at $90°$ at all.
CAll real numbers
This is the DOMAIN of $\arctan$ (every real $x$ has an arctangent), not its range. The range is bounded to the open interval $(-90°, 90°)$ — swapping domain and range is easy to do by accident once a function's inverse is involved.
D $0° \leq y \leq 180°$
This is the range of $\arccos$ , not $\arctan$ — the two inverse functions have different restricted ranges because $\cos\theta$ and $\tan\theta$ were restricted to different intervals in the first place.

Traps tested: Open interval treated as closed · Domain and range swapped · Arctan range confused with arccos

Question 71 mark

Which of these correctly rewrites $4\,\text{cosec}\,\theta$ as a single fraction in terms of $\sin\theta$ ?

$\frac{4}{\sin\theta}$
Correct. $\text{cosec}\,\theta = \frac{1}{\sin\theta}$ on its own, so $4\,\text{cosec}\,\theta = 4 \times \frac{1}{\sin\theta} = \frac{4}{\sin\theta}$ — the coefficient multiplies the finished reciprocal; it does not get absorbed into the denominator underneath $\sin\theta$ .
B $\frac{1}{4\sin\theta}$
This is a real, documented error on this exact spec point: a genuine WMA13 mark scheme names it directly, noting that writing "3cosecθ = 1/(3sinθ)" scores nothing for the identity mark unless the plain identity is also stated separately. Folding the coefficient into the denominator alongside $\sin\theta$ treats it as though it were inside the reciprocal from the start, when it multiplies the reciprocal once that reciprocal is already formed.
C $\frac{4}{\cos\theta}$
This reciprocates the wrong ratio — $\text{cosec}\,\theta$ is $\frac{1}{\sin\theta}$ , not $\frac{1}{\cos\theta}$ (that pairing belongs to $\sec\theta$ ). Getting the coefficient right doesn't help if the underlying reciprocal pairing is wrong.
D $\frac{1}{4}\sin\theta$
This neither reciprocates $\sin\theta$ nor keeps the coefficient in the right place — it is $\sin\theta$ scaled down by 4, when $\text{cosec}\,\theta$ needs $\sin\theta$ inverted first and then the result scaled up by 4.

Traps tested: Coefficient folded into denominator · Cosec computed as secant · Reciprocal step and coefficient both mishandled

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jun 2022 · Q7 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA13.

Secant, Cosecant, Cotangent and the Inverse Trigonometric Functions

Two things this lesson assumes — a recap, since neither has its own lesson yet

Three new names for three old ratios — and the one trap in the naming

Turning one true equation into another — by dividing every term by the same thing

Where sec²θ = 1 + tan²θ actually comes from

Why arcsin, arccos and arctan need restricted domains at all

The full solution set for sec²θ = 3 + tanθ, 0° ≤ θ < 360° — including the branch it's easy to stop after finding the nice one

Complete the chain — the full solution set for cosec²θ + cotθ = 3, 0° ≤ θ < 360°

Seven traps, one habit each — the checklist before the graded questions

Deep dive

Practice this for real