OCR A Level Mathematics A: Reciprocal and inverse trig, identities, compound angles, R-form and proof (H240) – 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 the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018), sections 1.05h to 1.05q. Questions 3 and 8 are Stage 1 content (shared with AS Level); the rest is Stage 2, A Level only. The numbers in square brackets in the answers show how many separate steps to aim to show. Learn the methods from the study guide or revision notes first. The course page is OCR A Level Mathematics, the printable list is the checklist, and the free diagnostics help you choose what to practise.

Questions

1. (a) Find the exact values of sec(2π/3), cosec(5π/4) and cot(π/6). [3] (b) State the range of y = 2 sec x, and the values of x in 0 ≤ x ≤ 2π at which it is undefined. [2]

2. (a) Find the exact values of arcsin(−√2/2) and arccos(−√3/2). [2] (b) Find the exact value of arcsin(sin 4π/3), explaining why it is not 4π/3. [2] (c) State the domain and range of y = arccos x and the coordinates of the end points of its graph. [2]

3. Solve 8cos²x + 2 sin x − 7 = 0 for 0 ≤ x < 360°, giving answers to 1 decimal place where necessary. [5]

4. (a) Show that 3 tan²θ + 5 sec θ + 1 = 0 can be written as 3 sec²θ + 5 sec θ − 2 = 0. [2] (b) Hence solve 3 tan²θ + 5 sec θ + 1 = 0 for 0 ≤ θ ≤ 2π, giving exact answers. [4]

5. A and B are angles with sin A = 5/13 (A acute) and cos B = −24/25 (B obtuse). (a) Find the exact value of cos(A + B). [3] (b) Find the exact value of tan(A − B). [3]

6. Show that tan 105° = −(2 + √3). [4]

7. Solve 3 sin 2x = 2 cos x for 0 ≤ x < 360°, giving answers to 1 decimal place where necessary. [5]

8. Solve tan(3θ − 20°) = 2 for −90° < θ < 90°, giving answers to 1 decimal place. [5]

9. f(θ) = 15 cos θ + 8 sin θ. (a) Express f(θ) in the form R cos(θ − α), where R > 0 and 0° < α < 90°. Give α to 2 decimal places. [3] (b) Solve f(θ) = 10 for 0 ≤ θ < 360°. [3] (c) Find the greatest value of 1/(25 + f(θ)) and the smallest positive value of θ at which it occurs. [3]

10. (a) Prove that cot x − tan x ≡ 2 cot 2x. [3] (b) Hence solve cot x − tan x = 2√3 for 0 < x < π, giving exact answers. [3]

11. In triangle XYZ, W is a point on XY such that ZW is perpendicular to XY. ZW = 12, XW = 5 and WY = 16. Angle XZW = A and angle WZY = B. (a) Write down tan A and tan B, and use a compound angle formula to show that tan(A + B) = 63/16. [3] (b) Find sin(A + B) exactly. Hence use the area of triangle XYZ, calculated in two ways, to check your value. [3]

12. A block of weight 40 N rests on rough horizontal ground. The coefficient of friction between the block and the ground is 0.75. A light rope attached to the block is pulled with tension P newtons at an angle θ above the horizontal, and the block is on the point of sliding. The friction force is 0.75R, where R newtons is the normal reaction. (a) By resolving vertically and horizontally, show that P(cos θ + 0.75 sin θ) = 30. [3] (b) Express cos θ + 0.75 sin θ in the form r cos(θ − α), giving r exactly and α in degrees to 2 decimal places. [3] (c) Find the least possible value of P and the value of θ at which it occurs. [2] (d) Find P when θ = 20°, giving your answer to 3 significant figures. [2]

Answers

1. (a) cos(2π/3) = −1/2, so sec(2π/3) = −2 [1]. sin(5π/4) = −√2/2, so cosec(5π/4) = −√2 [1]. tan(π/6) = 1/√3, so cot(π/6) = √3 [1] (b) Range: y ≤ −2 or y ≥ 2 [1]. Undefined where cos x = 0: x = π/2 and x = 3π/2 [1] Examiner insight: Take the sign from the quadrant before you invert; a correct size with the wrong sign is an easy slip on exact reciprocal values.

2. (a) arcsin(−√2/2) = −π/4 [1]. arccos(−√3/2) = 5π/6 [1] (b) sin(4π/3) = −√3/2 [1]. arcsin must return a value between −π/2 and π/2, so the answer is −π/3 [1] (c) Domain −1 ≤ x ≤ 1 and range 0 ≤ y ≤ π [1]. End points (−1, π) and (1, 0), with the graph decreasing [1] Examiner insight: For part (b), state the principal range as the reason; a reader cannot tell whether you know it from the number alone.

3. Replace cos²x by 1 − sin²x: 8(1 − sin²x) + 2 sin x − 7 = 0 [1]. This gives 8 sin²x − 2 sin x − 1 = 0 [1]. Factorise: (4 sin x + 1)(2 sin x − 1) = 0 [1]. sin x = 1/2 gives x = 30°, 150° [1]. sin x = −1/4 gives x = 194.5°, 345.5° [1] Examiner insight: Show the quadratic in sin x in full before factorising. Check that you have two solutions from each root, because sin x = −1/4 is negative and so lies in the third and fourth quadrants.

4. (a) Use tan²θ = sec²θ − 1: 3(sec²θ − 1) + 5 sec θ + 1 = 0 [1]. This simplifies to 3 sec²θ + 5 sec θ − 2 = 0, as required [1] (b) (3 sec θ − 1)(sec θ + 2) = 0 [1]. sec θ = 1/3 is impossible because |sec θ| ≥ 1 [1]. sec θ = −2 gives cos θ = −1/2 [1]. θ = 2π/3 and 4π/3 [1] Examiner insight: In a “show that” the last line should be the given result, written out. In (b), say why a root is rejected rather than quietly dropping it.

5. (a) cos A = 12/13 and sin B = 7/25 [1]. cos(A + B) = (12/13)(−24/25) − (5/13)(7/25) [1] = (−288 − 35)/325 = −323/325 [1] (b) tan A = 5/12 and tan B = −7/24 [1]. tan(A − B) = (5/12 + 7/24)/(1 + (5/12)(−7/24)) = (17/24)/(253/288) [1] = 204/253 [1] Examiner insight: Work out the missing ratios from a right-angled triangle first and note the sign from the quadrant. Leave exact fractions unconverted to decimals.

6. tan 105° = tan(60° + 45°) = (tan 60° + tan 45°)/(1 − tan 60° tan 45°) [1]. = (√3 + 1)/(1 − √3) [1]. Multiply top and bottom by (1 + √3): (√3 + 1)²/(1 − 3) [1]. = (4 + 2√3)/(−2) = −(2 + √3) [1] Examiner insight: Rationalise the denominator visibly. The target is given, so every line from the formula to the answer needs to appear.

7. 6 sin x cos x = 2 cos x [1]. cos x (6 sin x − 2) = 0 [1]. cos x = 0 gives x = 90°, 270° [1]. sin x = 1/3 [1] gives x = 19.5°, 160.5° [1] Examiner insight: Dividing both sides by cos x loses two of the four solutions. Move everything to one side and factorise.

8. If −90° < θ < 90°, then 3θ − 20° lies between −290° and 250° [1]. tan⁻¹ 2 = 63.43° [1]. Others: 63.43° − 180° = −116.57° and 63.43° + 180° = 243.43°; −296.57° is outside the interval [1]. Add 20° and divide by 3 [1]: θ = −32.2°, 27.8°, 87.8° [1] Examiner insight: Write the widened interval first, then list every angle in it. Stopping at the principal value leaves out two of the three answers.

9. (a) R cos α = 15 and R sin α = 8 [1]. R = √(225 + 64) = 17 [1]. tan α = 8/15, so α = 28.07° [1] (b) cos(θ − 28.07°) = 10/17 [1]. θ − 28.07° = ±53.97° [1]. θ = 82.0° or 334.1° [1] (c) The least value of f is −17, so the denominator has least value 25 − 17 = 8 and the greatest value is 1/8 [1]. This needs cos(θ − 28.07°) = −1 [1], so θ − 28.07° = 180° and θ = 208.1° [1] Examiner insight: For part (c), identify which end of f gives the extreme of the reciprocal before substituting; the greatest reciprocal comes from the smallest denominator.

10. (a) cot x − tan x = cos x/sin x − sin x/cos x = (cos²x − sin²x)/(sin x cos x) [1]. = cos 2x/(½ sin 2x) [1]. = 2 cos 2x/sin 2x = 2 cot 2x, as required [1] (b) 2 cot 2x = 2√3, so cot 2x = √3 and tan 2x = 1/√3 [1]. For 0 < 2x < 2π, 2x = π/6 or 7π/6 [1]. x = π/12, 7π/12 [1] Examiner insight: Start from one side only and name each identity you use. In (b), widen the interval for 2x before solving.

11. (a) tan A = 5/12 and tan B = 16/12 = 4/3 [1]. tan(A + B) = (5/12 + 4/3)/(1 − (5/12)(4/3)) [1] = (7/4)/(4/9) = 63/16 [1] (b) A right-angled triangle with sides 16 and 63 has hypotenuse 65, so sin(A + B) = 63/65 [1]. ZX = √(144 + 25) = 13 and ZY = √(144 + 256) = 20 [1]. Area = ½ × 13 × 20 × 63/65 = 126, and ½ × XY × ZW = ½ × 21 × 12 = 126, so the two agree [1] Examiner insight: Show the substitution into the formula before simplifying. For the check, write both areas as full expressions so the comparison is visible.

12. (a) Vertically: R + P sin θ = 40 [1]. Horizontally: P cos θ = 0.75R [1]. Substitute R = 40 − P sin θ: P cos θ = 30 − 0.75P sin θ, so P(cos θ + 0.75 sin θ) = 30 [1] (b) r cos α = 1 and r sin α = 0.75 [1]. r = √(1 + 0.5625) = 1.25 [1]. tan α = 0.75, so α = 36.87° [1] (c) P = 30/(1.25 cos(θ − 36.87°)) = 24/cos(θ − 36.87°) [1]. This is least when cos(θ − 36.87°) = 1, giving P = 24 N at θ = 36.87° [1] (d) cos 20° + 0.75 sin 20° = 1.1962 [1]. P = 30/1.1962 = 25.1 N [1] Examiner insight: Draw the force diagram and label both components before resolving. In (c), tie the minimum to the greatest value of the cosine, which links back to part (b).

Where marks are usually lost

• Giving the wrong sign on a reciprocal value because the quadrant was not checked first.
• Returning an angle outside the principal range for arcsin, arccos or arctan.
• Cancelling a trig factor instead of factorising, which removes solutions.
• Keeping a root such as sec θ = 1/3 that can never occur.
• Writing a plus sign in cos(A + B) where there should be a minus.
• Missing solutions because the interval was not widened for a multiple angle.
• Using a rounded value of α in later working, so the final answers drift.
• Starting a proof from both sides at once or skipping the line that reaches the target.
• Leaving out a force component when resolving, or using the wrong function for each.

Next steps

• Revise from the revision notes.
• Re-read the study guide.
• Course page: OCR A Level Mathematics and the checklist.
• Try the free diagnostics to find your weakest topics.
• Book a free trial class.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 1.05h to 1.05q.