OxfordAQA International AS and A-level Mathematics: Inverse trigonometric functions and secant, cosecant, cotangent (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow content area P2.4 (Trigonometry, first part) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level Unit P2, so it is A2 content and not examined at International AS. Angles are in radians unless a degree sign appears.

Links: course hub, checklist, study guide, revision notes and the free 10-minute diagnostics.

Questions

1. Given that tan θ = −7/24 and θ is obtuse, find the exact values of sec θ, cosec θ and cot θ. [3]

2. Find the exact value of each of the following, working by hand.

(a) sec(π/3) [1] (b) cosec(7π/6) [1] (c) cot(3π/4) [1]

3. Prove that tan x + cot x ≡ sec x cosec x. [3]

4. Solve 2tan²θ = 1 + sec θ for 0° ≤ θ ≤ 360°, giving answers to 1 decimal place where they are not exact. [5]

5. Solve cosec²x = 4 + 2cot x for 0 < x < 2π. Give exact answers where possible and otherwise answers to 3 decimal places. [4]

6. This question is about the graphs of reciprocal trigonometric functions.

(a) State the range of y = sec x. [1] (b) State the equations of the vertical asymptotes of y = sec x for 0 ≤ x ≤ 2π. [1] (c) Describe y = cosec x on the interval (0, 2π), giving its asymptote equations and the coordinates of its turning points. [2]

7. Find the exact value of each of the following.

(a) sin⁻¹(−√3/2) [1] (b) cos⁻¹(−1/2) [1] (c) tan⁻¹(−1) [1] (d) sin⁻¹(sin(4π/3)) [1]

8. The function f is defined by f(x) = 4sin⁻¹(x − 1) + π.

(a) State the domain of f. [1] (b) Find the range of f. [2] (c) Solve f(x) = 2π, giving your answer in exact form. [2]

9.

(a) Find the exact value of tan(cos⁻¹(−9/41)). [3] (b) Find the exact value of sin⁻¹(sin 2), where 2 is in radians. [2]

10. The graph of y = cos x for 0 ≤ x ≤ π is reflected to obtain the graph of y = cos⁻¹x.

(a) State the equation of the line of reflection. [1] (b) Write down the coordinates of the end points of y = cos⁻¹x. [2] (c) Explain why cos x must be restricted before an inverse can be defined, using the interval 0 ≤ x ≤ 2π as an example. [1] (d) State the equations of the two asymptotes of y = tan⁻¹x. [2]

11.

(a) Prove that 1/(1 + sin x) + 1/(1 − sin x) ≡ 2sec²x. [3] (b) Hence solve 1/(1 + sin x) + 1/(1 − sin x) = 8 for 0 ≤ x ≤ 2π. [3] (c) State the values of x in 0 ≤ x ≤ 2π for which the left-hand side in part (a) is undefined. [2]

12. The function h is defined by h(x) = tan⁻¹(2x) + π/4 for all real x.

(a) Find the range of h. [2] (b) Solve h(x) = π/2. [2] (c) Find an expression for h⁻¹(x) and state its domain. [3] (d) State the equations of the asymptotes of y = h(x). [1]

Answers

1. sec²θ = 1 + tan²θ = 1 + 49/576 = 625/576 [1]. θ is obtuse so cos θ is negative, giving sec θ = −25/24 [1]. cot θ = 1/tan θ = −24/7, and cosec²θ = 1 + cot²θ = 625/49, with cosec θ positive, so cosec θ = 25/7 [1]. sec θ = −25/24, cosec θ = 25/7, cot θ = −24/7 [3] Examiner insight: the sign is the part to defend. A written reason (“obtuse, so cosine is negative”) is more convincing than a square root left as plus or minus.

2. (a) cos(π/3) = 1/2, so sec(π/3) = 2 [1] (b) sin(7π/6) = −1/2, so cosec(7π/6) = −2 [1] (c) tan(3π/4) = −1, so cot(3π/4) = −1 [1] Examiner insight: write the sine, cosine or tangent value first and then take the reciprocal. Showing that intermediate value also lets you check the sign before you finish.

3. tan x + cot x = sin x/cos x + cos x/sin x [1]. Common denominator: (sin²x + cos²x)/(sin x cos x) [1]. This equals 1/(sin x cos x) = (1/cos x)(1/sin x) = sec x cosec x [1]. [3] Examiner insight: in a proof, every line should follow from the one before. Quoting sin²x + cos²x = 1 at the point you use it makes the reasoning easy to follow.

4. Use tan²θ = sec²θ − 1: 2sec²θ − 2 = 1 + sec θ [1]. So 2sec²θ − sec θ − 3 = 0 [1]. Factorise: (2sec θ − 3)(sec θ + 1) = 0 [1]. sec θ = 3/2 gives cos θ = 2/3, so θ = 48.2° or 311.8° [1]. sec θ = −1 gives cos θ = −1, so θ = 180° [1]. θ = 48.2°, 180°, 311.8° [5] Examiner insight: the value cos θ = −1 gives a single solution, and it is easy to drop. Check every root of the quadratic against the interval.

5. cosec²x = 1 + cot²x, so 1 + cot²x = 4 + 2cot x [1]. cot²x − 2cot x − 3 = 0, so (cot x − 3)(cot x + 1) = 0 [1]. cot x = 3 gives tan x = 1/3, so x = 0.322 or π + 0.322 = 3.463 [1]. cot x = −1 gives tan x = −1, so x = 3π/4 or 7π/4 [1]. x = 0.322, 3π/4, 3.463, 7π/4 [4] Examiner insight: convert cot to tan before using an inverse function. Keep extra decimal places until the final line, then round once.

6. (a) no values between −1 and 1: sec x ≤ −1 or sec x ≥ 1 [1] (b) cos x = 0 at π/2 and 3π/2 in this interval, so the asymptotes are x = π/2 and x = 3π/2 [1] (c) Vertical asymptotes at x = 0, x = π and x = 2π [1]. A branch above the axis with minimum point (π/2, 1), and a branch below the axis with maximum point (3π/2, −1) [1]. [2] Examiner insight: a sketch or description earns credit for named features, so state the asymptote equations and the turning-point coordinates explicitly. A curve that dips between y = −1 and y = 1 shows a misunderstanding.

7. (a) −π/3 [1] (b) angles with cosine −1/2 include 2π/3 and 4π/3, and only 2π/3 is in [0, π], so 2π/3 [1] (c) −π/4 [1] (d) sin(4π/3) = −√3/2, so the answer is sin⁻¹(−√3/2) = −π/3 (not 4π/3, which is outside the range) [1] Examiner insight: part (d) is a test of the range of sin⁻¹. An answer of 4π/3 suggests the function was treated as undoing the sine without a restriction.

8. (a) −1 ≤ x − 1 ≤ 1, so 0 ≤ x ≤ 2 [1] (b) −π/2 ≤ sin⁻¹(x − 1) ≤ π/2, so −2π ≤ 4sin⁻¹(x − 1) ≤ 2π [1]. Adding π: −π ≤ f(x) ≤ 3π [1] (c) 4sin⁻¹(x − 1) = π, so sin⁻¹(x − 1) = π/4 [1]. x − 1 = √2/2, so x = 1 + √2/2 [1] Examiner insight: for a range question, apply each operation to both end points in order. Adding π before multiplying by 4 is a common slip.

9. (a) Let α = cos⁻¹(−9/41). Then 0 ≤ α ≤ π and cos α < 0, so α is obtuse and sin α > 0 [1]. sin²α = 1 − 81/1681 = 1600/1681, so sin α = 40/41 [1]. tan α = (40/41)/(−9/41) = −40/9 [1] (b) 2 lies outside [−π/2, π/2], and sin 2 = sin(π − 2) [1]. π − 2 ≈ 1.14 lies inside the range, so the answer is π − 2 [1] Examiner insight: the answer 2 for part (b) is the typical wrong result. State why 2 is not an allowed output of sin⁻¹.

10. (a) y = x [1] (b) The end points of y = cos x are (0, 1) and (π, −1). Swapping coordinates: (1, 0) [1] and (−1, π) [1] (c) cos x takes some values more than once on 0 ≤ x ≤ 2π, for example cos(π/3) = cos(5π/3) = 1/2, so it is not one-to-one and has no inverse there [1] (d) y = π/2 [1] and y = −π/2 [1] Examiner insight: in part (c) a specific pair of inputs with the same output is a convincing explanation. A general statement that the graph repeats is thinner.

11. (a) Common denominator: [(1 − sin x) + (1 + sin x)]/[(1 + sin x)(1 − sin x)] [1]. This is 2/(1 − sin²x) [1]. Since 1 − sin²x = cos²x, it equals 2/cos²x = 2sec²x [1] (b) 2sec²x = 8, so sec²x = 4 and cos²x = 1/4 [1]. cos x = 1/2 or −1/2 [1]. x = π/3, 2π/3, 4π/3, 5π/3 [1] (c) The left-hand side is undefined when 1 + sin x = 0 or 1 − sin x = 0, so sin x = ±1 [1]. x = π/2 and x = 3π/2 [1] Examiner insight: in “hence” questions, use the result from part (a) directly. Starting again from the fractions is slower and invites errors.

12. (a) −π/2 < tan⁻¹(2x) < π/2 [1]. Adding π/4: −π/4 < h(x) < 3π/4 [1] (b) tan⁻¹(2x) = π/4 [1], so 2x = 1 and x = 1/2 [1] (c) Let y = tan⁻¹(2x) + π/4, so tan⁻¹(2x) = y − π/4 [1]. Then 2x = tan(y − π/4), so h⁻¹(x) = ½tan(x − π/4) [1]. The domain of h⁻¹ is the range of h: −π/4 < x < 3π/4 [1] (d) y = −π/4 and y = 3π/4 [1] Examiner insight: the domain of an inverse is the range of the original. State it as an inequality; a bare “all real x” for part (c) suggests the range from part (a) was ignored.

Where marks are usually lost

• Giving an angle outside the range of an inverse function, such as 4π/3 for sin⁻¹(sin(4π/3)).
• Taking the wrong sign for a square root without justifying it from the range or quadrant.
• Leaving out solutions, for example cos θ = −1 in question 4 or the second angle for each cot value in question 5.
• Using sec²x = 1 − tan²x or cosec²x = 1 − cot²x, with a sign error.
• Sketching sec and cosec curves with branches inside −1 < y < 1, or with no asymptotes marked.
• Proving an identity by working on both sides at once instead of starting from one side.
• Stating the domain of an inverse function as the domain of the original function.
• Forgetting that a strict inequality applies to tan⁻¹x, while sin⁻¹x and cos⁻¹x include their end points.
• Leaving answers in the wrong unit when the interval is given in degrees or in radians.

Next steps

• Condense the topic with the revision notes.
• Read the full explanations in the study guide.
• Return to the course hub and tick off the checklist.
• Take the free 10-minute diagnostics to find what to revise next.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: content area P2.4 Trigonometry (inverse trigonometric functions; secant, cosecant and cotangent).