Trigonometry Revision Notes — 9709 Paper 3

sec, cosec and cot

Syllabus requirement · §3.3

“

understand the relationship of the secant, cosecant and cotangent functions to cosine, sine and tangent, and use properties and graphs of all six trigonometric functions for angles of any magnitude.

”

Three new names, no new mathematics. Each is simply the reciprocal (1 divided by it) of one you already know:

\sec\theta = \frac{1}{\cos\theta} \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}

The pairing looks back to front: sec goes with cos, and cosec goes with sin. One way to remember it is the third letter: sec → cos, cosec → sin. Cot simply partners tan.

All three are undefined wherever the function underneath is zero, because you cannot divide by zero.

Each reciprocal graph drawn over its parent (dashed). Where the parent is zero, the reciprocal has a vertical asymptote: the zeros of cos become the asymptotes of sec, the zeros of sin those of cosec, and the zeros of tan those of cot.

Reading a reciprocal graph off the original

Where the original is zero, the reciprocal has a vertical asymptote: a vertical line the graph gets closer and closer to but never touches, because you cannot divide by zero.
Where the original is $\pm 1$ , so is the reciprocal, and the two curves touch.
Where the original is small, the reciprocal is large, and they share the same sign throughout.
So $\sec\theta$ and $\operatorname{cosec}\theta$ never take a value between $-1$ and $1$ : their range is $|y| \geqslant 1$ .

That last point is worth holding on to. An equation like $\sec\theta = 0.4$ has no solutions at all, and saying so is sometimes exactly what a question wants.

Function	Undefined (vertical asymptotes) where	Range	Repeats every
$\sec\theta = \dfrac{1}{\cos\theta}$	$\cos\theta = 0$ : $\theta = 90^\circ, 270^\circ, \ldots$	$y \leqslant -1$ or $y \geqslant 1$	$360^\circ$ ( $2\pi$ )
$\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$	$\sin\theta = 0$ : $\theta = 0^\circ, 180^\circ, 360^\circ, \ldots$	$y \leqslant -1$ or $y \geqslant 1$	$360^\circ$ ( $2\pi$ )
$\cot\theta = \dfrac{\cos\theta}{\sin\theta}$	$\sin\theta = 0$ : $\theta = 0^\circ, 180^\circ, 360^\circ, \ldots$	every real number	$180^\circ$ ( $\pi$ )

Each reciprocal repeats with the same period as its parent. cot θ is zero where cos θ is zero (90°, 270°, …), because the top of cos θ / sin θ is zero there.

Graphs with $2x$ inside. Exam sketches often use $\sec 2x$ or $\cot 2x$ rather than $\sec x$ . Doubling the angle squashes the graph sideways by a factor of $2$ , so everything happens twice as fast:

the period halves: $\sec 2x$ repeats every $180^\circ$ ( $\pi$ ), and $\cot 2x$ every $90^\circ$ ( $\tfrac12\pi$ );
the asymptotes move to where the inside makes the parent zero. For $\sec 2x$ that is $\cos 2x = 0$ , so $2x = \tfrac12\pi, \tfrac32\pi, \ldots$ and $x = \tfrac14\pi, \tfrac34\pi, \ldots$ . For $\cot 2x$ it is $\sin 2x = 0$ , so $x = 0, \tfrac12\pi, \pi, \ldots$

The routine is always: find where the parent of the inside angle is zero, then solve for $x$ .

Values at any angle

Your calculator has no $\sec$ , $\operatorname{cosec}$ or $\cot$ button, and it does not need one. To find $\sec\theta$ , find $\cos\theta$ first and then take $1$ divided by it. The same goes for $\operatorname{cosec}$ (via $\sin$ ) and $\cot$ (via $\tan$ ).

This also settles the sign. A number and its reciprocal always have the same sign, so $\sec\theta$ is negative exactly where $\cos\theta$ is negative (the second and third quadrants), $\operatorname{cosec}\theta$ is negative exactly where $\sin\theta$ is, and $\cot\theta$ is negative exactly where $\tan\theta$ is.

Exact values of the reciprocal functions

Without a calculator, find the exact values of (a) $\sec 120^\circ$ (b) $\operatorname{cosec} 225^\circ$ (c) $\cot(-60^\circ)$ .

Show full working

1
(a) Write it as a reciprocal: $\sec 120^\circ = \frac{1}{\cos 120^\circ}$
2
$120^\circ$ is in the second quadrant, where $\cos$ is negative. Its related acute angle is $180^\circ - 120^\circ = 60^\circ$ . So $\cos 120^\circ = -\cos 60^\circ = -\tfrac12$
The acute angle gives the size of the value; the quadrant gives its sign. This is the Paper 1 routine, unchanged.
3
Take the reciprocal. Dividing $1$ by $-\tfrac12$ gives $\sec 120^\circ = \frac{1}{-\tfrac12} = -2$
4
(b) $\operatorname{cosec} 225^\circ = \dfrac{1}{\sin 225^\circ}$ . The angle $225^\circ$ is in the third quadrant, where $\sin$ is negative, and its related acute angle is $225^\circ - 180^\circ = 45^\circ$ . So $\sin 225^\circ = -\sin 45^\circ = -\frac{1}{\sqrt2}$
5
Take the reciprocal: $\operatorname{cosec} 225^\circ = \frac{1}{-\tfrac{1}{\sqrt2}} = -\sqrt2$
6
(c) Write $\cot$ as cos over sin: $\cot(-60^\circ) = \frac{\cos(-60^\circ)}{\sin(-60^\circ)}$
7
A negative angle is measured clockwise, into the fourth quadrant, where $\cos$ is positive and $\sin$ is negative: $\cos(-60^\circ) = \tfrac12, \qquad \sin(-60^\circ) = -\tfrac{\sqrt3}{2}$
8
Divide: $\cot(-60^\circ) = \frac{\tfrac12}{-\tfrac{\sqrt3}{2}} = -\frac{1}{\sqrt3}$
The halves cancel. −1/√3 and −√3/3 are the same number; either is exact.

Answer

(a) $-2$ (b) $-\sqrt2$ (c) $-\dfrac{1}{\sqrt3}$

Recap: finding every solution

Every equation in this note ends with a line like $\cos\theta = -0.3$ or $\tan 2x = \sqrt3$ , so here is the Paper 1 routine in one place.

Flip a reciprocal function first. $\sec\theta = k$ means $\cos\theta = \tfrac1k$ ; $\operatorname{cosec}\theta = k$ means $\sin\theta = \tfrac1k$ ; $\cot\theta = k$ means $\tan\theta = \tfrac1k$ .
Get one angle $\alpha$ from the calculator ( $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ ).
Get the partner angle in the same cycle from the symmetry of the graph:
- $\sin\theta = k$ : $\;\alpha$ and $180^\circ - \alpha$ (in radians, $\pi - \alpha$ );
- $\cos\theta = k$ : $\;\alpha$ and $360^\circ - \alpha$ (in radians, $2\pi - \alpha$ );
- $\tan\theta = k$ : $\;\alpha$ and $\alpha + 180^\circ$ (in radians, $\alpha + \pi$ ).
Add or subtract a full period ( $360^\circ$ or $2\pi$ for sin and cos; $180^\circ$ or $\pi$ for tan) until you have covered the whole interval, and throw away anything outside it.
If the angle inside is not plain $\theta$ (say $2\theta$ ), change the interval to match first: $0^\circ < \theta < 180^\circ$ becomes $0^\circ < 2\theta < 360^\circ$ . Find every value of $2\theta$ in that interval, then halve each one. The Double angles section shows this in detail.

Each horizontal line cuts one cycle twice. For sine the second angle is 180° − α; for cosine it is 360° − α; for tangent it is α + 180°.

Solving with sec and cot

Solve, for $0^\circ \leqslant \theta \leqslant 360^\circ$ , (a) $3\sec\theta + 6 = 0$ (b) $\cot\theta = 2$ .

Show full working

1
(a) Make $\sec\theta$ the subject. Subtract $6$ from both sides, then divide by $3$ : $3\sec\theta = -6 \quad\Longrightarrow\quad \sec\theta = -2$
2
Flip to cosine. $\sec\theta = -2$ means $\dfrac{1}{\cos\theta} = -2$ , so $\cos\theta = -\tfrac12$
Check the range first: −2 is not strictly between −1 and 1, so sec θ = −2 has solutions. After flipping, −½ is between −1 and 1, as a cosine must be.
3
The calculator gives $\cos^{-1}\!\left(-\tfrac12\right) = 120^\circ$ .
4
The cosine partner is $360^\circ - 120^\circ = 240^\circ$ . Adding or subtracting $360^\circ$ leaves the interval, so there are no more.
5
(b) Flip to tangent: $\cot\theta = 2$ means $\tan\theta = \tfrac12$
6
The calculator gives $\tan^{-1}(0.5) = 26.57^\circ$ , which is $26.6^\circ$ to 1 d.p.
7
The tangent partner is $26.57^\circ + 180^\circ = 206.57^\circ$ . Adding another $180^\circ$ gives $386.6^\circ$ , outside the interval.

Answer

(a) $\theta = 120^\circ,\ 240^\circ$ (b) $\theta = 26.6^\circ,\ 206.6^\circ$

When in doubt, convert to sin and cos

Almost every trig problem simplifies if you rewrite everything in terms of $\sin$ and $\cos$ . It is rarely the shortest route, but it is the most reliable one, and mark schemes accept it as an alternative method. Use it whenever you cannot see which identity the question wants.

Sketching a pair of graphs to count roots

Some Paper 3 questions say "by sketching a suitable pair of graphs, show that the equation … has exactly one root", and one of the two graphs is often $\sec$ , $\operatorname{cosec}$ or $\cot$ . You need these graphs exactly right: where the asymptotes are, and which way each branch curves.

The argument: split the equation into two curves and sketch both. If one curve only goes up while the other only goes down, they can cross at most once. If one curve is above the other at the left end of the interval and below it at the right end, they must cross at least once. Together that is exactly one root. The Numerical Solution of Equations note covers this method in full.

The two marks go to: the first graph drawn correctly (asymptotes, intercepts, shape), and the second graph plus the crossing point marked with a dot and a sentence saying it is the root.

Reading mark-scheme codes

The past-paper solutions in this note mention mark codes. M is a method mark (a correct method, even with an arithmetic slip). A is an accuracy mark for a correct answer, and needs the M mark before it. B is a mark for a correct result on its own. A misread (for example answering in degrees when radians were asked) usually costs one mark.

Sketching sec and cot to show one root

9709/32 O/N 2024 Q2(a)2 marks

By sketching a suitable pair of graphs, show that the equation $\cot 2x = \sec x$ has exactly one root in the interval $0 < x < \tfrac12\pi$ .

Show full working

The mark scheme's sketch: y = cot 2x and y = sec x on 0 < x < ½π, meeting exactly once. The smaller panel shows an accepted alternative, y = tan 2x against y = cos x.

1
Split the equation into two curves: $y = \cot 2x$ and $y = \sec x$ , both for $0 < x < \tfrac12\pi$ .
2
Asymptotes of $\cot 2x$ : where $\sin 2x = 0$ , i.e. $2x = 0, \pi$ , so $x = 0$ and $x = \tfrac12\pi$ . These are the two ends of the interval.
This is the 'find where the parent of the inside angle is zero' routine from the text above.
3
Shape of $\cot 2x$ between them: it falls from $+\infty$ (just after $x = 0$ ) to $-\infty$ (just before $x = \tfrac12\pi$ ), crossing zero where $\cos 2x = 0$ , at $x = \tfrac14\pi$ . It is decreasing all the way.
4
Shape of $\sec x$ : $\sec 0 = 1$ , and as $x \to \tfrac12\pi$ , $\cos x \to 0^+$ , so $\sec x \to +\infty$ . It is increasing all the way, starting at $1$ .
One curve only falls and the other only rises, so they can cross at most once.
5
Compare the ends. Near $x = 0$ , $\cot 2x$ is huge and positive, above $\sec x \approx 1$ . Near $x = \tfrac12\pi$ , $\cot 2x \to -\infty$ while $\sec x \to +\infty$ , so $\sec x$ is above. The curves swap order, so they cross at least once.
6
At most once and at least once means exactly one root. Draw both curves, mark the crossing with a dot, and state that its $x$ -coordinate is the root.

Answer

The sketch shows exactly one intersection of $y = \cot 2x$ and $y = \sec x$ in $0 < x < \tfrac12\pi$ , so the equation has exactly one root there.

Your turn

Range, exact values and the features of a sec 2x graph first, then solving, then two exam questions: one converted to sin and cos, one answered with a sketch of sec 2x.

1
State the range of $\sec\theta$ and of $\operatorname{cosec}\theta$ . Hence say, without solving, whether each equation has any solutions: (a) $\sec\theta = 0.6$ (b) $\operatorname{cosec}\theta = -2.5$ (c) $\operatorname{cosec}\theta = 1$ .
Stuck? Show hint
sec and cosec never take a value strictly between −1 and 1. Decide whether each number is in that gap.
Show solution
1. 1
  $\sec\theta$ and $\operatorname{cosec}\theta$ never take a value strictly between $-1$ and $1$ , so their range is $|y| \geqslant 1$ .
2. 2
  (a) $0.6$ lies strictly between $-1$ and $1$ , so $\sec\theta = 0.6$ has no solutions.
3. 3
  (b) $-2.5$ satisfies $|-2.5| \geqslant 1$ , so it is in range — $\operatorname{cosec}\theta = -2.5$ does have solutions.
4. 4
  (c) $1$ is on the boundary of the range, $|1| = 1$ , so it is attainable — $\operatorname{cosec}\theta = 1$ does have a solution (at $\theta = 90^\circ$ ).
  The boundary case is the one people miss — |y| ⩾ 1 includes the endpoints, it is not a strict inequality.
Answer
(a) no solutions (b) solutions exist (c) solutions exist
2
Find the exact values of (a) $\operatorname{cosec} 300^\circ$ (b) $\cot 135^\circ$ .
Stuck? Show hint
Find sin 300° and tan 135° first, using the related acute angle and the quadrant, then take reciprocals.
Show solution
1. 1
  (a) $\operatorname{cosec} 300^\circ = \dfrac{1}{\sin 300^\circ}$ . The angle $300^\circ$ is in the fourth quadrant, where $\sin$ is negative, and its related acute angle is $360^\circ - 300^\circ = 60^\circ$ .
2. 2
  So $\sin 300^\circ = -\sin 60^\circ = -\dfrac{\sqrt3}{2}$ .
3. 3
  Take the reciprocal: $\operatorname{cosec} 300^\circ = \frac{1}{-\tfrac{\sqrt3}{2}} = -\frac{2}{\sqrt3}$
  Dividing by a fraction means multiplying by its reciprocal: 1 ÷ (√3/2) = 2/√3.
4. 4
  (b) $\cot 135^\circ = \dfrac{1}{\tan 135^\circ}$ . The angle $135^\circ$ is in the second quadrant, where $\tan$ is negative, and its related acute angle is $45^\circ$ .
5. 5
  So $\tan 135^\circ = -\tan 45^\circ = -1$ , and $\cot 135^\circ = \frac{1}{-1} = -1$
Answer
(a) $-\dfrac{2}{\sqrt3}$ (or $-\dfrac{2\sqrt3}{3}$ ) (b) $-1$
3
For $y = \sec 2x$ with $0 \leqslant x \leqslant \pi$ , state (a) the equations of the asymptotes (b) the period (c) the points where the graph turns (d) the range. Then sketch the graph.
Stuck? Show hint
Start from y = cos 2x: where is it 0, where is it 1, where is it −1?
Show solution
y = sec 2x drawn over its parent y = cos 2x (dashed). The asymptotes are where cos 2x = 0.
1. 1
  (a) Asymptotes where $\cos 2x = 0$ : $2x = \tfrac12\pi$ or $\tfrac32\pi$ , so $x = \tfrac14\pi$ and $x = \tfrac34\pi$ .
2. 2
  (b) $\sec\theta$ repeats every $2\pi$ , so $\sec 2x$ repeats when $2x$ goes up by $2\pi$ , i.e. every $\pi$ in $x$ .
3. 3
  (c) Where $\cos 2x = \pm 1$ , $\sec 2x$ is also $\pm 1$ and the curve turns. $\cos 2x = 1$ at $x = 0$ and $x = \pi$ , giving $(0, 1)$ and $(\pi, 1)$ (bottoms of U-shapes). $\cos 2x = -1$ at $x = \tfrac12\pi$ , giving $(\tfrac12\pi, -1)$ (the top of an upside-down U).
4. 4
  (d) As for $\sec x$ : $y \leqslant -1$ or $y \geqslant 1$ .
5. 5
  Sketch (see the figure): a half U rising from $(0, 1)$ towards the asymptote $x = \tfrac14\pi$ ; an upside-down U between $x = \tfrac14\pi$ and $x = \tfrac34\pi$ with its top at $(\tfrac12\pi, -1)$ ; a half U coming down from the asymptote $x = \tfrac34\pi$ to $(\pi, 1)$ . It is the sec graph squashed sideways by a factor of 2.
Answer
(a) $x = \tfrac14\pi,\ \tfrac34\pi$ (b) $\pi$ (c) $(0, 1)$ , $(\tfrac12\pi, -1)$ , $(\pi, 1)$ (d) $y \leqslant -1$ or $y \geqslant 1$
4
Solve $2\operatorname{cosec}\theta = 5$ for $0 < \theta < 2\pi$ , giving your answers correct to 3 significant figures.
Stuck? Show hint
Make cosec θ the subject, flip it to sin θ, and remember the interval is in radians.
Show solution
1. 1
  Divide by $2$ : $\operatorname{cosec}\theta = 2.5$ .
2. 2
  Flip to sine: $\sin\theta = \frac{1}{2.5} = 0.4$
3. 3
  With the calculator in radians, $\sin^{-1}(0.4) = 0.4115$ .
  The interval 0 < θ < 2π is in radians, so the answers must be too.
4. 4
  The sine partner is $\pi - 0.4115 = 2.7301$ . Both lie between $0$ and $2\pi$ ; adding $2\pi$ to either goes past $2\pi$ .
Answer
$\theta = 0.412,\ 2.73$
59709/31 O/N 2016 Q35 marks
Express the equation $\sec\theta = 3\cos\theta + \tan\theta$ as a quadratic equation in $\sin\theta$ . Hence solve this equation for $-90^\circ < \theta < 90^\circ$ .
Stuck? Show hint
Rewrite every term in sin and cos, multiply through by cos θ, then use sin²θ + cos²θ ≡ 1 to remove the remaining cos²θ.
Show solution
1. 1
  Rewrite in sin and cos: $\frac{1}{\cos\theta} = 3\cos\theta + \frac{\sin\theta}{\cos\theta}$
2. 2
  Multiply every term by $\cos\theta$ to clear the fractions: $1 = 3\cos^2\theta + \sin\theta$
3. 3
  The question wants a quadratic in $\sin\theta$ , so replace $\cos^2\theta$ by $1-\sin^2\theta$ : $1 = 3\left(1-\sin^2\theta\right) + \sin\theta$
4. 4
  Expand the bracket: $1 = 3 - 3\sin^2\theta + \sin\theta$
5. 5
  Move every term to the left, so the $\sin^2\theta$ term is positive: $3\sin^2\theta - \sin\theta - 2 = 0$
6. 6
  Factorise, writing $s = \sin\theta$ . Numbers multiplying to $3 \times (-2) = -6$ and adding to $-1$ are $-3$ and $2$ , so split the middle term: $3s^2 - 3s + 2s - 2 = 3s(s - 1) + 2(s - 1) = (3s + 2)(s - 1)$ So $(3\sin\theta+2)(\sin\theta-1)=0$ .
7. 7
  So $\sin\theta = 1$ or $\sin\theta = -\tfrac23$ .
8. 8
  Reject $\sin\theta=1$ . It gives $\theta = 90^\circ$ , which is the excluded endpoint of $-90^\circ < \theta < 90^\circ$ .
  A strict inequality excludes its own endpoints — always check whether a 'nice' root like this one actually lies inside the open interval.
9. 9
  $\sin\theta = -\tfrac23$ gives $\theta = -41.8^\circ$ , which does lie in the interval.
Answer
$\theta = -41.8^\circ$ only
69709/31 O/N 2025 Q9(a)2 marks
By sketching a suitable pair of graphs, show that the equation $\sec 2x = -\mathrm{e}^x$ has only one root in the interval $0 < x < \tfrac12\pi$ .
Stuck? Show hint
Find where cos 2x = 0 for the asymptote. Which branch of sec 2x is negative, like −eˣ?
Show solution
The mark scheme's sketch of y = sec 2x and y = −eˣ on 0 < x < ½π.
1. 1
  Asymptote of $\sec 2x$ : where $\cos 2x = 0$ , i.e. $2x = \tfrac12\pi$ , so $x = \tfrac14\pi$ .
2. 2
  Left branch, $0 < x < \tfrac14\pi$ : $\cos 2x$ runs from $1$ down to $0^+$ , so $\sec 2x$ runs from $1$ up to $+\infty$ . It is positive here.
3. 3
  Right branch, $\tfrac14\pi < x < \tfrac12\pi$ : $\cos 2x$ runs from $0^-$ down to $-1$ , so $\sec 2x$ rises from $-\infty$ up to $-1$ . It is negative and increasing here.
4. 4
  $y = -\mathrm{e}^x$ starts at $-1$ when $x = 0$ and decreases (to $-\mathrm{e}^{\pi/2} \approx -4.81$ ). It is always negative.
5. 5
  On the left branch, $\sec 2x > 0 > -\mathrm{e}^x$ , so there is no crossing there.
  A positive curve and a negative curve can never meet.
6. 6
  On the right branch, $\sec 2x$ starts below $-\mathrm{e}^x$ (it comes up from $-\infty$ ) and ends above it ( $-1 > -4.81$ ). One curve rises and the other falls, so they cross exactly once. Mark it with a dot.
Answer
The sketch shows one intersection, on the branch of $y = \sec 2x$ between $\tfrac14\pi$ and $\tfrac12\pi$ , so there is only one root.

Practise sec, cosec and cotReal past-paper questions · Secant, cosecant and cotangent functions and their graphs

Checking your access…

Can you do all of these?

Write sec, cosec and cot in terms of cos, sin and tan without hesitating
Find exact values of sec, cosec and cot at any angle, with the sign from the quadrant
Find every solution in an interval using the sin, cos and tan partner rules
Sketch all six graphs, and say where the asymptotes of sec, cosec and cot are
State the asymptotes, period and range of graphs such as y = sec 2x and y = cot 2x
Find the greatest or least value of 1/(k + R sin(θ + α)) by making the bottom as small or large as possible
Split an equation involving sec, cosec or cot into two curves and use their shape to prove exactly one root
Derive sec²θ ≡ 1 + tan²θ by dividing, rather than recalling it
Use the rearranged forms sec²θ − tan²θ ≡ 1 to spot a difference of two squares
Expand sin(A ± B), cos(A ± B), tan(A ± B), getting the cos sign flip right
Find exact values such as sin 75° and tan 15° by splitting the angle
Collapse a compound-angle equation to a single tan by dividing by cos x
Given one trig ratio and the quadrant, find the others and the double angle values exactly
Factorise out a common cos θ or sin θ instead of dividing by it
Choose the version of cos 2A that leaves one function in the equation
Apply a double angle formula to any pair where one angle is twice the other
Stretch the interval for 2θ, and shrink it for ½x, before listing solutions
Express a sin θ + b cos θ in all four R-forms by comparing coefficients
Expand a compound angle first when the expression is not yet a sin θ + b cos θ
Adjust the interval for the whole bracket (e.g. 2θ − α) before solving an R-form equation
Read the maximum, minimum and their positions straight off an R-form
Remember that the least value of a squared R-form is 0, not −R²
Prove an identity working on one side only, ending at the printed expression
Reject roots outside the range of the function before writing answers down
Answer in the units the interval was stated in
Add two compound expansions to turn a product of trig functions into a sum
Substitute a new angle into a proved identity when the “hence” part demands it
Halve (or otherwise convert) every answer back to the variable actually asked for
Reject an impossible root such as tan² = −1 before listing solutions

Now do the questions

109 real Paper 3 parts from 2021–2025, sorted by difficulty, with mark schemes

Logarithmic and Exponential Functions

Differentiation

Trigonometry

sec, cosec and cot

Values at any angle

Recap: finding every solution

The Pythagorean identities

Compound angles

Can you do all of these?

sec, cosec and cot

Values at any angle

Recap: finding every solution

The rest of this note

Can you do all of these?