plumiSec, cosec, cot and inverse trig functions: AQA A‑Level Maths questions
AQA A-Level Maths past-paper questions on E.5 Sec, cosec, cot and inverse trig functions, 2018–2025, with mark schemes and examiner’s reports where available. Add MadAsMaths questions to your practice, or browse them separately by topic and difficulty.
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All 14 questions on Sec, cosec, cot and inverse trig functions, as a list
- June 2025, Paper 2, Q2 (1 mark): One of the diagrams below shows the graph of y = cosec x° for 0 ≤ x ≤ 360 Identify the correct graph. Tick (✓) one box. [1 mark] x y x y 1 –1 1 –1 180 360 90 27
- June 2025, Paper 3, Q2 (1 mark): The diagram shows the graph of y = arctan x x a – a y O The graph has horizontal asymptotes at y = –a and y = a State the value of a Circle your answer. [1 mark
- June 2024, Paper 1, Q4 (1 mark): One of the diagrams below shows the graph of y = arccos x Identify the graph of y = arccos x Tick (✓) one box. [1 mark] y x O π 2 –1 y x O 1 π 2
- June 2023, Paper 1, Q13 (9 marks): The function f is defined by f (x) = arccosx for 0 × x × a The curve with equation y = f (x) is shown below. O a x 2 2 y 13 (a) State the value of a [1 mark] 13
- June 2023, Paper 2, Q8 (10 marks): (a) Given that cos y 6= ≤1 , prove the identity 1 1 − cos y + 1 1 + cos y ≥ 2 cosec2 y [4 marks] 8 (b) Hence, find the set of values of A for which the equation
- June 2022, Paper 1, Q7 (3 marks): Sketch the graph of y = cot x × p 2 for 0 − x − 2p [3 marks] x y O 2
- June 2022, Paper 1, Q15 (16 marks): (a) Given that y = cosecy 15 (a) (i) Express y in terms of siny. [1 mark] 15 (a) (ii) Hence, prove that dy dy = × cosecy cot y [3 marks] 15 (a) (iii) Show that
- November 2021, Paper 1, Q8 (9 marks): (a) Given that 9 sin2 y + sin 2y = 8 show that 8 cot2 y − 2 cot y − 1 = 0 [4 marks] 8 (b) Hence, solve 9 sin2 y + sin 2y = 8 in the interval 0 < y < 2p Give you
- November 2021, Paper 3, Q1 (1 mark): The graph of y = arccosx is shown. P x y State the coordinates of the end point P. Circle your answer. [1 mark] (−p, 1) (−1, p) − p 2 , 1 −1, p 2
- November 2020, Paper 2, Q2 (1 mark): Which one of the following equations has no real solutions? Tick (3) one box. [1 mark] cot x = 0 ln x = 0 jx + 1j = 0 sec x = 0
- June 2019, Paper 1, Q7 (11 marks): (a) By sketching the graphs of y = 1 x and y = sec 2x on the axes below, show that the equation 1 x = sec 2x has exactly one solution for x > 0 [3 marks] x ϖ 2ϖ
- June 2019, Paper 1, Q12 (7 marks): (a) Show that the equation 2 cot2 x + 2 cosec2 x = 1 + 4 cosecx can be written in the form a cosec2 x + b cosecx + c = 0 [2 marks] 12 (b) Hence, given x is obtu
- June 2019, Paper 3, Q1 (1 mark): f (x) = arcsinx State the maximum possible domain of f Tick (3) one box. [1 mark] fx 2 R : −1 × x × 1g n x 2 R : − p 2 × x × p 2 o fx 2 R : −p × x × pg fx 2 R :
- June 2018, Paper 1, Q12 (10 marks): p(x) = 30x3 − 7x2 − 7x + 2 12 (a) Prove that (2x + 1) is a factor of p(x) [2 marks] 12 (b) Factorise p(x) completely. [3 marks] 12 (c) Prove that there are no r
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