AQA A-Level Mathematics Paper 2, June 2025: Question 8
6 marks · Medium difficulty · Multi-step Problem
Use trigonometric identities to evaluate sec²(x) - tan²(x), simplify an expression involving sec(x) and tan(x) to an integer constant, and state values of x where the expression is undefined.
Practise this questionQuestion
![Question 8 has three parts. Part (a) states: Given that cos x ≠ 0, state the value of sec² x - tan² x [1 mark]. Part (b) states: Show that (3 sec x + 5 tan x)(5 sec x - 3 tan x) - (16 tan x) / (cos x) = N where N is an integer to be found [4 marks]. Part (c) states: State the two values of x between 0° and 360° for which the value of N in part (b) is not valid [1 mark].](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2Fquestions%252FAQA-ALEVEL-MAT.Q25.L2.08.SRC.aqa-7357-2%25257C2025%25257CJUN%25257Cuntiered-1790799441160.png%3Falt%3Dmedia&w=1000)
Question text
8 (a) Given that cos x ≠ 0 state the value of sec2 x – tan2 x
[1 mark]
8 (b) Show that
x + x x – x – 16tan x N
(3sec 5tan )(5sec 3tan ) =
cos x
where N is an integer to be found.
[4 marks]
8 (c) State the two values of x between 0° and 360° for which the value of N in part (b) is
not valid.
[1 mark]
(12) …
Mark scheme
Show the mark scheme
Q Marking instructions AO Marks Typical solution
8(a) States the value 1 1.2 B1 1
Subtotal 1
8(b) Expands the given brackets to 3.1a M1
obtain at least three of the 16 tan x
(3 sec x + 5 tan x)(5 sec x − 3 tan x) −
following four terms cos x
15 sec2 x
= 15sec2 x − 9sec x tan x + 25sec x tan x
25sec x tan x
OE 2
−9sec x tan x −15tan x −16sec x tan x
−15 tan2 x
Expands the given brackets to 1.1b A1
obtain ( 2 2 )
22 =15 sec x − tan x
15 sec x + 25 sec x tan x − 9 sec x tan x − 15 tan x
or = 15
15 sec2 x +16 sec x tan x −15 tan2 x
OE
Evaluates 2.2a B1
15 sec2 x −15 tan2 x to 15
Or
Evaluates
16 tan x
16sec x tan x − to 0
cos x
Completes reasoned argument to 2.1 R1
obtain 15
Argument must include a
conversion of 16sec x tan x or
16tan x
to a common form before
cos x
cancellation
CSO
Subtotal 4
8(c) Obtains 90 and 270 2.2a R1 x = 90, x = 270
CAO
Subtotal 1
Question 8 Total 6
How to answer it
Trigonometric Identities & Validity of Reciprocal Functions
This question assesses your fluency with Year 2 trigonometric identities, algebraic manipulation of reciprocal functions, and the domains over which trigonometric functions are defined.
- Pythagorean Identities: Recalling and rearranging 1 + tan² x ≡ sec² x .
- Reciprocal Definitions: Applying the definition sec x = 1 / cos x to simplify algebraic fractions.
- Mathematical Rigour: Providing a fully reasoned proof (CSO) without jumping steps or making unjustified cancellations.
- Undefined Values: Identifying angles where trigonometric functions are invalid due to division by zero ( cos x = 0 ).
State the value of sec² x − tan² x
Target: AO1.2 (Recall and basic manipulation)
✅ Correct Answer
1
💡 Key Knowledge
From the fundamental identity sin² x + cos² x ≡ 1 , divide every term by cos² x :
tan² x + 1 ≡ sec² x
Rearranging this directly gives:
sec² x − tan² x = 1
Show that (3 sec x + 5 tan x)(5 sec x − 3 tan x) − (16 tan x / cos x) = N
Target: AO3.1a (1), AO1.1b (1), AO2.2a (1), AO2.1 (1)
📐 Step-by-Step Calculation
- Expand the double brackets:
(3 sec x + 5 tan x)(5 sec x − 3 tan x)
= 15 sec² x − 9 sec x tan x + 25 sec x tan x − 15 tan² x
= 15 sec² x + 16 sec x tan x − 15 tan² xM1: Expanding brackets to get at least 3 correct terms.
A1: All 4 terms expanded and collected correctly to obtain 15 sec² x + 16 sec x tan x − 15 tan² x . - Substitute into the full expression:
15 sec² x + 16 sec x tan x − 15 tan² x − (16 tan x / cos x) - Convert to a common form before cancelling:
Recognise that 1 / cos x = sec x , so:
16 tan x / cos x = 16 sec x tan x
The expression becomes:
15 sec² x + 16 sec x tan x − 15 tan² x − 16 sec x tan x
= 15 sec² x − 15 tan² xB1: Recognising that 16 sec x tan x − (16 tan x / cos x) = 0 , OR evaluating 15 sec² x − 15 tan² x = 15 . - Apply the identity from part (a):
15(sec² x − tan² x) = 15(1) = 15
Hence, N = 15.R1: Fully completed reasoned argument obtaining 15. The conversion of 16 sec x tan x or 16 tan x / cos x to a common form must be explicitly written before cancelling (CSO).
🧠 Exam Technique: Securing the Reasoning Mark (R1)
This is a "Show that" question. The mark scheme explicitly demands CSO (Correct Solution Only) with full working:
- You cannot simply cross out 16 sec x tan x and 16 tan x / cos x without explicitly writing either 16 sec x tan x or 16 tan x / cos x in the same form.
- Factor out 15 clearly: write 15(sec² x − tan² x) = 15(1) = 15 to make your use of part (a) obvious.
❌ Common Errors
- Sign slips during FOIL expansion: Calculating 5 tan x × (−3 tan x) as +15 tan² x .
- Premature cancellation: Crossing out the fractional term without establishing that 1 / cos x ≡ sec x .
- Forgetting to state N: Always conclude clearly with N = 15 .
State the two values of x between 0° and 360° for which the value of N is not valid
Target: AO2.2a (Analysis and deduction)
✅ Correct Answer
x = 90° and x = 270°
💡 Mathematical Explanation
The original question states: cos x ≠ 0 .
Both tan x = sin x / cos x and sec x = 1 / cos x are undefined when the denominator is zero:
cos x = 0
Solving for 0° < x < 360° :
x = 90°, 270°
At these angles, the functions have asymptotes, so the algebraic identity does not hold.
❌ Common Misconceptions
- Giving radians instead of degrees: Writing π/2, 3π/2 when the question explicitly specified the range in degrees ( 0° and 360° ).
- Setting N = 0 or tan x = 0: Incorrectly solving sin x = 0 , leading to wrong angles like 180° .
- Listing only one angle: Missing 270° in the given range.
Topics
Pure Mathematics · E: Trigonometry
Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.