June 2025 Paper 2 Q8

AQACurrent spec6 marksProof Trigonometry

(a) Given that \(\cos x \neq 0\) state the value of \(\sec^2 x - \tan^2 x\) [1 mark]

(b) Show that\[(3\sec x + 5\tan x)(5\sec x - 3\tan x) - \frac{16\tan x}{\cos x} = N\]where \(N\) is an integer to be found. [4 marks]

(c) State the two values of \(x\) between \(0^\circ\) and \(360^\circ\) for which the value of \(N\) in part (b) is not valid. [1 mark]

Scheme	Marks	AO
States the value 1	B1	1.2
	(1)

Mark scheme (b)

Scheme	Marks	AO
Expands the given brackets to obtain at least three of the following four terms \(15\sec^2 x\) \(25\sec x\tan x\) \(-9\sec x\tan x\) \(-15\tan^2 x\) OE	M1	3.1a
Expands the given brackets to obtain \(15\sec^2 x + 25\sec x\tan x - 9\sec x\tan x - 15\tan^2 x\) or \(15\sec^2 x + 16\sec x\tan x - 15\tan^2 x\) OE	A1	1.1b
Evaluates \(15\sec^2 x - 15\tan^2 x\) to 15 Or Evaluates \(16\sec x\tan x - \dfrac{16\tan x}{\cos x}\) to 0	B1	2.2a
Completes reasoned argument to obtain 15 Argument must include a conversion of \(16\sec x\tan x\) or \(\dfrac{16\tan x}{\cos x}\) to a common form before cancellation CSO	R1	2.1
	(4)

Typical solution

\[(3\sec x + 5\tan x)(5\sec x - 3\tan x) - \frac{16\tan x}{\cos x}\]\[= 15\sec^2 x - 9\sec x\tan x + 25\sec x\tan x - 15\tan^2 x - 16\sec x\tan x\]\[= 15\left(\sec^2 x - \tan^2 x\right)\]\[= 15\]

Scheme	Marks	AO
Obtains 90 and 270 CAO	R1	2.2a
	(1)
	(6 marks)