Edexcel International A Level Maths · P3 · Pure Mathematics 3
Reciprocal and inverse trigonometric functions
28 free, high-quality past-paper-style questions on Reciprocal and inverse trigonometric functions for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 234 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-3-B-01P3 · A2Easy5 marks
This question is about the values of the reciprocal trigonometric functions. Answer parts (a) to (c) without a calculator, showing your working.
(a)
Find the exact value of cosec(4π/3), giving your answer with a rational denominator.
(2)
(b)
Find the exact value of sec 135°.
(1)
(c)
Find the exact value of cot(7π/4).
(1)
(d)
Find the value of cosec 2.5, giving your answer to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
cosec(4π/3) = 1/sin(4π/3) = 1/(−√3/2)
M1
= −2/√3 = −2√3/3
A1
Part (b)
1/cos 135° = 1/(−√2/2) = −√2
B1
Part (c)
1/tan(7π/4) = 1/(−1) = −1
B1
Part (d)
1/sin 2.5 = 1/0.5985 = 1.67 (radian mode)
B1
Common error: Using degree mode for cosec 2.5. With no ° symbol the angle is in radians; in degree mode you would get 1/sin 2.5° ≈ 22.9.
Give each answer to this question as an exact value in radians or as an exact number. The inverse trigonometric functions arcsin, arccos and arctan take their usual ranges.
(a)
Write down the value of arcsin(½).
(1)
(b)
Write down the value of arccos(−√3/2).
(1)
(c)
Write down the value of arctan(−√3).
(1)
(d)
Find the exact value of sin(arccos(−5/13)).
(2)
(e)
Find the value of arccos(cos(4π/3)).
(1)
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Answer
Mark
Notes
Part (a)
π/6
B1
Part (b)
5π/6
B1
Part (c)
−π/3
B1
Part (d)
let α = arccos(−5/13); 0 ≤ α ≤ π so sin α ≥ 0 and sin α = √(1 − 25/169)
M1
= 12/13
A1
Part (e)
cos(4π/3) = −½, so arccos(−½) = 2π/3
B1
Exam tip: An inverse trig function gives exactly one angle, in its own range: [−π/2, π/2] for arcsin, [0, π] for arccos, (−π/2, π/2) for arctan. That is why arccos(cos(4π/3)) is not 4π/3.
The angle θ is reflex with 270° < θ < 360°, and sec θ = 5/2. Do not use a calculator in this question.
(a)
Use an appropriate identity to find the exact value of tan θ.
(3)
(b)
Write down the exact value of cot θ.
(1)
(c)
Use the identity cosec2θ = 1 + cot2θ to find the exact value of cosec θ.
(2)
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Answer
Mark
Notes
Part (a)
tan2θ = sec2θ − 1
M1
= 25/4 − 1 = 21/4
A1
θ is in the fourth quadrant, so tan θ < 0: tan θ = −√21/2
A1
Part (b)
cot θ = −2/√21 = −2√21/21 (ft their tan θ)
B1
Part (c)
cosec2θ = 1 + 4/21 = 25/21
M1
sin θ < 0, so cosec θ = −5/√21 = −5√21/21
A1
Exam tip: The identity gives a square, so it always leaves a ± to settle. In the fourth quadrant cos (and sec) is positive while sin, tan, cosec and cot are negative.
Figure 1 shows the curve with equation y = cosec x, for 0° < x < 360°, x ≠ 180°. The curve has a minimum point A and a maximum point B. The line y = 2.5 meets the curve at P and Q, and the line y = −2.5 meets the curve at R and S.
(a)
Write down the coordinates of A and of B.
(2)
(b)
Find the x-coordinates of P and Q, giving your answers to 1 decimal place.
(3)
(c)
Use the symmetry of the curve, or otherwise, to find the x-coordinates of R and S, giving your answers to 1 decimal place.
(2)
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Answer
Mark
Notes
Part (a)
A(90°, 1)
B1
B(270°, −1)
B1
Part (b)
cosec x = 2.5 ⇒ sin x = 0.4
M1
P: x = 23.6°
A1
Q: x = 180° − 23.6° = 156.4°
A1
Part (c)
sin x = −0.4: x = 180° + 23.6°
M1
R: 203.6°, S: 336.4°
A1
Exam tip: Turn every cosec equation into a sine equation first: cosec x = 2.5 means sin x = 0.4. The figure then shows you how many solutions to expect in each branch.
Figure 1 shows the right-angled triangle ABC, with the lengths of its sides in centimetres. The angle ACB is θ. Do not use a calculator in this question.
(a)
Find the exact values of sec θ, cosec θ and cot θ.
(3)
(b)
The obtuse angle φ is such that φ = 180° − θ. Find the exact values of sec φ and cot φ.
(2)
(c)
Use your answers to part (a) to verify that cosec2θ = 1 + cot2θ.
(1)
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Answer
Mark
Notes
Part (a)
sec θ = 1/cos θ = 17/15
B1
cosec θ = 1/sin θ = 17/8
B1
cot θ = 1/tan θ = 15/8
B1
Part (b)
cos φ = −cos θ, so sec φ = −17/15
B1
tan φ = −tan θ, so cot φ = −15/8
B1
Part (c)
cosec2θ = 289/64 and 1 + cot2θ = 1 + 225/64 = 289/64, so they are equal
B1
Common error: Taking sec θ as hypotenuse/opposite. sec goes with cos, so it is hypotenuse/adjacent; cosec goes with sin (hypotenuse/opposite).
Figure 1 shows the curves y = arcsin x and y = arccos x, for −1 ≤ x ≤ 1, where arcsin and arccos take their usual ranges. The curves meet at the point P. The line x = −0.5 meets y = arcsin x at Q and y = arccos x at R.
(a)
State the range of y = arcsin x and the range of y = arccos x.
(2)
(b)
Find the exact coordinates of P.
(2)
(c)
Find the exact length of QR.
(2)
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Answer
Mark
Notes
Part (a)
arcsin: −π/2 ≤ y ≤ π/2
B1
arccos: 0 ≤ y ≤ π
B1
Part (b)
at P, sin y = cos y = x with 0 ≤ y ≤ π/2, so tan y = 1 and y = π/4
M1
P(√2/2, π/4)
A1
Part (c)
Q: arcsin(−0.5) = −π/6 and R: arccos(−0.5) = 2π/3
M1
QR = 2π/3 − (−π/6) = 5π/6
A1
Exam tip: arcsin of a negative number is negative (range −π/2 to π/2), but arccos of a negative number lies between π/2 and π. Read the signs off the figure before you subtract.
Figure 1 shows the curves y = sec2x and y = tan2x, for 0 ≤ x < π/2. The points A on y = tan2x and B on y = sec2x both have x-coordinate π/3. The line y = 5 meets y = sec2x at C and y = tan2x at D.
(a)
Find the exact y-coordinates of A and B, and use an identity to explain why the vertical distance between the two curves is the same for every value of x in the interval.
(3)
(b)
Find the x-coordinates of C and D, giving your answers to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
A: tan2(π/3) = (√3)2 = 3
B1
B: sec2(π/3) = 22 = 4
B1
sec2x − tan2x = 1 for all x, so the upper curve is always exactly 1 unit above the lower one
B1
Part (b)
C: sec2x = 5 ⇒ tan2x = 4
M1
tan x = 2 (positive in the interval), x = 1.11
A1
D: tan x = √5, x = 1.15
A1
Exam tip: sec² x = 5 is quickest through the identity: tan² x = 5 − 1 = 4. Only the positive root of tan x is needed here because 0 ≤ x < π/2.
The curve C has equation y = 1 + 3 cosec x, for −π < x < π, x ≠ 0.
(a)
Write down the equation of the asymptote of C in this interval.
(1)
(b)
Find the coordinates of the two turning points of C, stating which is a maximum and which is a minimum.
(2)
(c)
Find the set of values of k for which the equation 1 + 3 cosec x = k has no solutions in the interval.
(2)
(d)
Solve 1 + 3 cosec x = 7 in the interval, giving exact answers.
(3)
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Answer
Mark
Notes
Part (a)
x = 0
B1
Part (b)
minimum (π/2, 4)
B1
maximum (−π/2, −2)
B1
Part (c)
the curve has y ≤ −2 or y ≥ 4
M1
−2 < k < 4
A1
Part (d)
cosec x = 2, so sin x = ½
M1
x = π/6
A1
x = 5π/6 (and no negative solutions, since sin x < 0 there)
A1
Model sketch
Figure: y = 1 + 3 cosec x for −π < x < π, x ≠ 0
Exam tip: Transform the key features of cosec x: the asymptotes stay where sin x = 0, and the turning points (π/2, 1) and (−π/2, −1) go to (π/2, 1 + 3) and (−π/2, 1 − 3).
In this question, x is not a multiple of π/2, so that every expression is defined.
(a)
Prove that (1 + cot2x)(sec2x − 1) ≡ sec2x.
(3)
(b)
Prove that (sec x − cos x)/tan x ≡ sin x.
(3)
(c)
Hence solve (sec x − cos x)/tan x = 0.3 for 0 < x < 2π, giving your answers to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
LHS = cosec2x × tan2x (both identities)
M1
= (1/sin2x)(sin2x/cos2x)
M1
= 1/cos2x = sec2x = RHS
A1
Part (b)
sec x − cos x = (1 − cos2x)/cos x
M1
= sin2x/cos x
M1
dividing by sin x/cos x gives sin2x/sin x = sin x = RHS
A1
Part (c)
sin x = 0.3
M1
x = 0.305
A1
x = π − 0.3047 = 2.84
A1
Exam tip: In a proof, work from one side only and end with “= RHS”. Writing sec, cosec, tan and cot in terms of sin and cos is almost always a safe first move.
Let y = arcsin x, where −1 < x < 1 and arcsin takes its usual range.
(a)
Explain why cos y ≥ 0, and hence show that cos y = √(1 − x2).
(2)
(b)
Hence write down an expression for tan y in terms of x.
(1)
(c)
Find the exact value of tan(arcsin(2/3)), giving your answer with a rational denominator.
(2)
(d)
Use the identity sec2β = 1 + tan2β to find the exact value of cos(arctan(√7/3)).
(3)
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Answer
Mark
Notes
Part (a)
−π/2 ≤ y ≤ π/2, where cosine is not negative
B1
sin y = x, so cos2y = 1 − x2 and cos y = +√(1 − x2)
B1
Part (b)
tan y = x/√(1 − x2)
B1
Part (c)
(2/3)/√(1 − 4/9) = (2/3)/(√5/3)
M1
= 2/√5 = 2√5/5
A1
Part (d)
sec2β = 1 + 7/9 = 16/9 where β = arctan(√7/3)
M1
−π/2 < β < π/2 so cos β > 0 and sec β = 4/3
M1
cos(arctan(√7/3)) = 3/4
A1
Model sketch
Figure: right-angled triangle with sin y = x; not to scale
Exam tip: To find a trig function of an inverse trig value, name the angle, write down the one ratio you know, and use the range of the inverse function to fix the sign of the ratio you want.
Each part of this question involves a reciprocal trigonometric function of a transformed angle.
(a)
Solve cosec(2x − π/3) = −√2 for 0 ≤ x ≤ π, giving your answers as exact multiples of π.
(5)
(b)
Solve cot(θ/2) = 1.5 for 0° ≤ θ < 360°, giving your answer to 1 decimal place.
(3)
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Answer
Mark
Notes
Part (a)
sin(2x − π/3) = −1/√2
M1
−π/3 ≤ 2x − π/3 ≤ 5π/3
M1
2x − π/3 = −π/4 or 5π/4 (7π/4 is outside the interval)
A1
x = π/24
A1
x = 19π/24
A1
Part (b)
tan(θ/2) = 2/3
M1
θ/2 = 33.69° (and 0° ≤ θ/2 < 180° allows no other value)
A1
θ = 67.4° only
A1
Common error: Solving for the angle inside the bracket on the original interval. Change the interval first: here 2x − π/3 runs from −π/3 to 5π/3, so −π/4 counts and 7π/4 does not.
Figure 1 shows part of the curve with equation y = a cot(x − c), for 0 < x < 2π, where a and c are constants with a > 0 and 0 < c < π/2. The curve passes through P(5π/12, 2) and crosses the x-axis at Q(2π/3, 0).
(a)
Use the point Q to find the value of c.
(2)
(b)
Use the point P to find the value of a.
(2)
(c)
Write down the equations of the two vertical asymptotes shown in Figure 1.
(2)
(d)
Solve a cot(x − c) = 5 for 0 < x < 2π, giving your answers to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
cot(2π/3 − c) = 0 ⇒ 2π/3 − c = π/2 (the only option with 0 < c < π/2)
M1
c = π/6
A1
Part (b)
a cot(5π/12 − π/6) = a cot(π/4) = 2
M1
a = 2
A1
Part (c)
cot(x − π/6) is undefined where x − π/6 = 0 or π
M1
x = π/6 and x = 7π/6
A1
Part (d)
cot(x − π/6) = 2.5 ⇒ tan(x − π/6) = 0.4
M1
x − π/6 = 0.3805 or 0.3805 + π, so x = 0.904
A1
x = 4.05
A1
Exam tip: The zeros of cot u are at u = π/2 + nπ and its asymptotes at u = nπ. Shift these by c to match the x-intercept and asymptotes on the graph.
Figure 1 shows the curve with equation y = a arctan(x − 1) + b, where a and b are positive constants and arctan takes its usual range. The curve has horizontal asymptotes y = 3π/2 and y = −π/2.
(a)
Find the value of a and the value of b.
(3)
(b)
Show that the curve passes through the origin.
(1)
(c)
Write down the coordinates of the point where the curve crosses the line y = π/2.
(1)
(d)
Solve a arctan(x − 1) + b = π, giving the exact answer.
(2)
(e)
Find the x-coordinate of the point on the curve where y = 1, giving your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
b + aπ/2 = 3π/2 and b − aπ/2 = −π/2 (as arctan → ±π/2)
M1
b = π/2
A1
a = 2
A1
Part (b)
x = 0: 2 arctan(−1) + π/2 = 2(−π/4) + π/2 = 0
B1
Part (c)
arctan(x − 1) = 0 ⇒ (1, π/2)
B1
Part (d)
arctan(x − 1) = π/4
M1
x − 1 = 1, x = 2
A1
Part (e)
arctan(x − 1) = (1 − π/2)/2 = −0.2854
M1
x = 1 + tan(−0.2854) = 0.707
A1
Exam tip: arctan tends to π/2 and −π/2, so a arctan(...) + b has asymptotes y = b ± aπ/2. Halfway between the asymptotes gives b; half the gap gives aπ/2.
Figure 1 shows the plan view of a spotlight S which is 4 m from a straight wall. The point N on the wall is the point nearest to S, so that SN is perpendicular to the wall. The beam of light makes an angle θ with SN, where 0° ≤ θ < 90°, and meets the wall at B. The distance SB = d metres and NB = s metres.
(a)
Show that d = 4 sec θ.
(1)
(b)
Find the value of θ for which d = 6.5, giving your answer in degrees to 1 decimal place.
(2)
(c)
Write s in terms of θ, and use the identity sec2θ = 1 + tan2θ to show that d2 = 16 + s2.
(3)
(d)
Find the value of θ for which d = 2s.
(2)
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Answer
Mark
Notes
Part (a)
cos θ = 4/d, so d = 4/cos θ = 4 sec θ
B1
Part (b)
sec θ = 6.5/4 ⇒ cos θ = 4/6.5
M1
θ = 52.0°
A1
Part (c)
s = 4 tan θ
B1
d2 = 16 sec2θ = 16(1 + tan2θ)
M1
= 16 + 16 tan2θ = 16 + s2
A1
Part (d)
4 sec θ = 8 tan θ ⇒ 1/cos θ = 2 sin θ/cos θ ⇒ sin θ = ½
M1
θ = 30°
A1
Common error: Writing d = 4 cos θ. In triangle SNB the hypotenuse is d, so cos θ = 4/d and d = 4/cos θ = 4 sec θ. Check: d must be at least 4, and sec θ ≥ 1 on this interval.
Figure 1 shows the curve y = sin x, for −2π ≤ x ≤ 2π. Two sections of the curve are highlighted: S1, for −π/2 ≤ x ≤ π/2, and S2, for π/2 ≤ x ≤ 3π/2.
(a)
Explain why the function sin x, with domain −2π ≤ x ≤ 2π, does not have an inverse function.
(1)
(b)
The function f is sin x with the domain of section S1, so that f−1(x) = arcsin x. State the domain and the range of f−1.
(2)
(c)
The function g is sin x with the domain of section S2. Show that g−1(x) = π − arcsin x.
(2)
(d)
Find the exact value of g−1(0.5), and solve g−1(x) = 4π/3.
(2)
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Answer
Mark
Notes
Part (a)
it is many-to-one: for example the line y = ½ meets the curve more than once, so the inverse would not be a function
B1
Part (b)
domain −1 ≤ x ≤ 1
B1
range −π/2 ≤ f−1(x) ≤ π/2
B1
Part (c)
let y = g−1(x), so sin y = x with π/2 ≤ y ≤ 3π/2; then sin(π − y) = sin y = x
M1
−π/2 ≤ π − y ≤ π/2, the range of arcsin, so π − y = arcsin x and y = π − arcsin x
A1
Part (d)
g−1(0.5) = π − π/6 = 5π/6
B1
x = sin(4π/3) = −√3/2
B1
Exam tip: A trig function only has an inverse on a section where it is one-to-one. Any such section works, but arcsin is defined using the section −π/2 ≤ x ≤ π/2.
Figure 1 shows the curves y = 3 cot x and y = 2 cosec2x − 3, for 0° < x < 360°, x ≠ 180°. The curves meet at the four points A, B, C and D.
(a)
Show that the x-coordinates of the points of intersection satisfy 2 cot2x − 3 cot x − 1 = 0.
(2)
(b)
Hence find the x-coordinates of A, B, C and D, giving your answers to 1 decimal place.
(4)
(c)
Explain why no point of the curve y = 2 cosec2x − 3 lies below the line y = −1, and write down the coordinates of its minimum points.
(2)
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Answer
Mark
Notes
Part (a)
2(1 + cot2x) − 3 = 3 cot x
M1
2 cot2x − 3 cot x − 1 = 0
A1
Part (b)
cot x = (3 ± √17)/4
M1
cot x = 1.7808 or −0.2808, so tan x = 0.5616 or −3.5616
A1
tan x = 0.5616: x = 29.3°, 209.3°
A1
tan x = −3.5616: x = 105.7°, 285.7°
A1
Part (c)
cosec2x ≥ 1, so 2 cosec2x − 3 ≥ −1
B1
(90°, −1) and (270°, −1)
B1
Exam tip: A cosec² term next to cot is the signal for cosec² x = 1 + cot² x. Use the quadratic formula when it does not factorise, and keep 4 s.f. for cot x before taking the inverse.
The function f is defined by f(x) = sec x, 0 ≤ x < π/2. Figure 1 shows the curves y = f(x) and y = f−1(x), together with the line y = x. The point P(2, π/3) lies on y = f−1(x).
(a)
State the domain and the range of f−1.
(2)
(b)
Show that f−1(x) = arccos(1/x).
(2)
(c)
Verify that P lies on y = f−1(x), and write down the coordinates of the point on y = f(x) that corresponds to P.
(2)
(d)
Find f−1(5), giving your answer to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
domain x ≥ 1 (the range of f)
B1
range 0 ≤ f−1(x) < π/2
B1
Part (b)
y = f−1(x) ⇒ sec y = x, so cos y = 1/x
M1
0 ≤ y < π/2 lies in the range of arccos, so y = arccos(1/x)
A1
Part (c)
arccos(1/2) = π/3
B1
(π/3, 2) (reflection in y = x)
B1
Part (d)
arccos(0.2) = 1.37
B1
Exam tip: The domain of an inverse function is the range of the original function. From the graph, sec x on 0 ≤ x < π/2 takes every value from 1 upwards.
The curve C has equation y = a + b sec x, for 0 ≤ x ≤ 2π, x ≠ π/2, x ≠ 3π/2, where a and b are constants with b > 0. The points (0, 1) and (π/3, 5) lie on C.
(a)
Find the value of a and the value of b.
(3)
(b)
Write down the equations of the asymptotes of C and find the coordinates of its maximum point.
(3)
(c)
Solve the equation a + b sec x = 13 in the given interval, giving your answers to 3 significant figures.
(3)
(d)
Find the set of values of k for which the equation a + b sec x = k has exactly two solutions in the given interval.
(2)
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Answer
Mark
Notes
Part (a)
a + b = 1 (sec 0 = 1)
M1
a + 2b = 5 (sec(π/3) = 2)
M1
a = −3, b = 4
A1
Part (b)
x = π/2 and x = 3π/2
B1
maximum where sec x = −1, at x = π
M1
(π, −7) (ft their a, b)
A1
Part (c)
sec x = 4, so cos x = ¼
M1
x = 1.32
A1
x = 2π − 1.318 = 4.97
A1
Part (d)
considers the outer branches (y ≥ 1, including both end points at k = 1) and the middle branch (y ≤ −7, one solution only at k = −7)
M1
k < −7 or k ≥ 1
A1
Model sketch
Figure: y = −3 + 4 sec x for 0 ≤ x ≤ 2π
Exam tip: Count solutions with the sketch in front of you. Check the special values of k separately: k = 1 meets the curve at both end points, while k = −7 touches it only at the maximum.
In this question, 0 < x < 2π and x is not a multiple of π/2.
(a)
Prove that cosec x / (cosec x − sin x) ≡ sec2x.
(4)
(b)
Hence solve cosec x / (cosec x − sin x) = 3 tan x + 5, giving exact answers where possible and other answers to 3 significant figures.
(6)
(c)
Explain why the identity in part (a) is not valid when x = π/2.
(1)
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Answer
Mark
Notes
Part (a)
LHS = (1/sin x) / (1/sin x − sin x)
M1
= (1/sin x) / ((1 − sin2x)/sin x) = 1/(1 − sin2x)
M1
= 1/cos2x
M1
= sec2x = RHS
A1
Part (b)
sec2x = 3 tan x + 5 and sec2x = 1 + tan2x
M1
tan2x − 3 tan x − 4 = 0
A1
(tan x − 4)(tan x + 1) = 0
M1
tan x = 4 or tan x = −1
A1
x = 3π/4, 7π/4
A1
x = 1.33, 4.47
A1
Part (c)
cosec(π/2) − sin(π/2) = 1 − 1 = 0, so the left-hand side would need division by zero (and sec(π/2) is undefined)
B1
Exam tip: “Hence” means use part (a): replace the awkward fraction by sec2x, then the identity sec2x = 1 + tan2x turns the equation into a quadratic in tan x.
This question is about the equation 3 tan2α − 2 sec α = 5 and a transformed version of it.
(a)
Show that 3 tan2α − 2 sec α = 5 can be written as 3 sec2α − 2 sec α − 8 = 0.
(3)
(b)
Hence solve 3 tan2(2θ − 30°) − 2 sec(2θ − 30°) = 5 for 0° ≤ θ ≤ 180°, giving your answers to 1 decimal place where appropriate.
(8)
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Answer
Mark
Notes
Part (a)
tan2α = sec2α − 1
M1
3 sec2α − 3 − 2 sec α = 5
M1
3 sec2α − 2 sec α − 8 = 0
A1
Part (b)
(3 sec α + 4)(sec α − 2) = 0
M1
sec α = 2 or −4/3
A1
cos α = ½ or cos α = −¾
M1
α = 2θ − 30° with −30° ≤ α ≤ 330°
M1
cos α = ½: α = 60°, 300° (−60° is outside)
A1
cos α = −¾: α = 138.59°, 221.41°
A1
θ = (α + 30°)/2: 45°, 165°
A1
84.3°, 125.7°
A1
Common error: Missing or adding solutions after the substitution α = 2θ − 30°. Find the interval for α first (−30° to 330°), list every α in it, and only then convert back to θ.
For x > 0, let α = arctan x, where arctan takes its usual range.
(a)
State the range of possible values of α, and show that cot α = 1/x.
(2)
(b)
Hence prove that arctan x + arctan(1/x) = π/2 for x > 0.
(3)
(c)
Hence solve arctan x − arctan(1/x) = π/6 for x > 0.
(3)
(d)
State the value of arctan x + arctan(1/x) when x < 0.
(1)
(e)
Use the identity cosec2α = 1 + cot2α to find the exact value of cosec(arctan(3/4)).
(2)
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Answer
Mark
Notes
Part (a)
0 < α < π/2
B1
tan α = x, so cot α = 1/tan α = 1/x
B1
Part (b)
tan(π/2 − α) = cot α = 1/x (or: in a right-angled triangle with shorter sides x and 1 the two acute angles are α and π/2 − α)
M1
0 < π/2 − α < π/2, which lies in the range of arctan, so arctan(1/x) = π/2 − α
M1
arctan x + arctan(1/x) = α + π/2 − α = π/2, as required
A1
Part (c)
arctan x − (π/2 − arctan x) = π/6
M1
2 arctan x = 2π/3, so arctan x = π/3
A1
x = √3
A1
Part (d)
−π/2 (arctan is an odd function)
B1
Part (e)
cosec2α = 1 + (4/3)2 = 25/9
M1
0 < α < π/2 so cosec α > 0: 5/3
A1
Model sketch
Figure: right-angled triangle with tan α = x; the other acute angle is π/2 − α; not to scale
Exam tip: Whenever you claim that arctan(something) equals an angle, check that the angle lies in (−π/2, π/2). That one line is what turns a calculation into a proof.
Figure 1 shows part of a circle with centre O and radius 1. The point P lies on the circle, where angle POT = θ and 45° < θ < 90°. The tangent to the circle at P meets the x-axis at T and the y-axis at S.
(a)
Explain why angle OPT = 90°, and show that OT = sec θ.
(2)
(b)
Show that OS = cosec θ and PS = cot θ.
(3)
(c)
Use Pythagoras' theorem in triangle OPS to prove that cosec2θ = 1 + cot2θ.
(1)
(d)
Given that ST = 13/6, find θ, giving your answer in degrees to 1 decimal place, and find the exact length OT.
(4)
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Answer
Mark
Notes
Part (a)
a tangent is perpendicular to the radius at the point of contact
B1
in triangle OPT, cos θ = OP/OT = 1/OT, so OT = sec θ
B1
Part (b)
angle POS = 90° − θ, and angle OPS = 90°
M1
OS = 1/cos(90° − θ) = 1/sin θ = cosec θ
A1
PS = tan(90° − θ) = cos θ/sin θ = cot θ
A1
Part (c)
OS2 = OP2 + PS2 gives cosec2θ = 1 + cot2θ
B1
Part (d)
PT = tan θ, so tan θ + cot θ = 13/6 ⇒ 6 tan2θ − 13 tan θ + 6 = 0
M1
(3 tan θ − 2)(2 tan θ − 3) = 0; θ > 45° so tan θ > 1 and tan θ = 3/2
A1
θ = 56.3°
A1
OT = sec θ = √(1 + 9/4) = √13/2
A1
Exam tip: In a figure built on a unit circle, each length is one ratio: look for the right angle, then write the side you want over the side of length 1.
Figure 1 shows the curve with equation y = sec x + tan x, for −π/2 < x < π/2. The line y = 3 meets the curve at the point Q. As x → −π/2, y → 0, and as x → π/2, y → ∞.
(a)
Prove that (sec x + tan x)(sec x − tan x) ≡ 1.
(2)
(b)
Given that sec x − tan x = 1/3, use part (a) to show that sec x + tan x = 3, and hence find the exact values of sec x and tan x.
(4)
(c)
Hence find the x-coordinate of Q, giving your answer to 3 significant figures.
(1)
(d)
Use Figure 1 to state the range of values of sec x + tan x for −π/2 < x < π/2. Hence find the range of values of sec x − tan x on the same interval.
(3)
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Answer
Mark
Notes
Part (a)
LHS = sec2x − tan2x
M1
= (1 + tan2x) − tan2x = 1 = RHS
A1
Part (b)
sec x + tan x = 1 ÷ (1/3) = 3
B1
adding the two equations: 2 sec x = 10/3
M1
sec x = 5/3
A1
tan x = 3 − 5/3 = 4/3
A1
Part (c)
cos x = 3/5 with tan x > 0, so x = 0.927
B1
Part (d)
sec x + tan x > 0, taking every positive value
B1
sec x − tan x = 1/(sec x + tan x) by part (a)
M1
so sec x − tan x also takes every positive value: sec x − tan x > 0
A1
Exam tip: sec² x − tan² x = 1 factorises as a difference of two squares. Knowing one factor immediately gives the other as its reciprocal.
Figure 1 shows the curve with equation y = 2 sec2θ − 5 tan θ, for 0° ≤ θ ≤ 180°, θ ≠ 90°. The curve crosses the θ-axis at A and B and has a minimum point M. The end points of the curve are (0°, 2) and (180°, 2).
(a)
Show that the equation of the curve can be written as y = 2 tan2θ − 5 tan θ + 2.
(2)
(b)
Find the θ-coordinates of A and B, giving your answers to 1 decimal place.
(3)
(c)
By completing the square, find the exact y-coordinate of M and its θ-coordinate to 1 decimal place.
(3)
(d)
Find the set of values of k for which the equation 2 sec2θ − 5 tan θ = k has exactly two solutions for 0° ≤ θ ≤ 180°, θ ≠ 90°.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
sec2θ = 1 + tan2θ
M1
y = 2 + 2 tan2θ − 5 tan θ = 2 tan2θ − 5 tan θ + 2
A1
Part (b)
(2 tan θ − 1)(tan θ − 2) = 0
M1
tan θ = ½: θ = 26.6°
A1
tan θ = 2: θ = 63.4°
A1
Part (c)
y = 2(tan θ − 5/4)2 − 9/8
M1
minimum y = −9/8
A1
tan θ = 5/4, θ = 51.3°
A1
Part (d)
for 90° < θ ≤ 180°, tan θ ≤ 0 so y ≥ 2: this branch gives one solution for each k ≥ 2 and none for k < 2
M1
k = 2: tan θ = 0 or 5/2, giving θ = 0°, 180° and 68.2°, three solutions, so k = 2 is excluded
M1
k > 2: one solution on each branch; −9/8 < k < 2: two solutions on the left branch only
M1
−9/8 < k < 2 or k > 2
A1
Common error: Forgetting that the end points belong to the curve. At k = 2 the line meets the curve at both (0°, 2) and (180°, 2) as well as on the left branch, so k = 2 gives three solutions, not two.
Exam tip: An inequality between two curves is read from the sketch: find where they cross, then also mark the asymptotes, because a curve can jump from below to above the other at an asymptote without any crossing.
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