Question 4 - A-Level Maths

OCR MEI C4 — Question 4 7 marks

Exam Board	OCR MEI
Module	C4 (Core Mathematics 4)
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Prove identity then solve equation
Difficulty	Standard +0.3 This is a two-part question requiring a standard double-angle identity proof followed by a trigonometric equation. The proof uses the known formula for tan 2θ and reciprocal relationships (routine manipulation). The equation solving requires algebraic rearrangement leading to a quadratic in tan θ, which is a standard C4 technique. Slightly above average due to the multi-step nature and need to find all solutions in the given range, but still a typical textbook exercise.
Spec	1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs 1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals

4 Show that $\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }$.
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$

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Question 4:

Answer	Marks	Guidance
Answer	Mark	Guidance
$\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$	M1	oe, e.g. converting either side into a one-line fraction(s) involving $\sin\theta$ and $\cos\theta$
$\cot 2\theta = \dfrac{1}{\tan 2\theta} = \dfrac{1-\tan^2\theta}{2\tan\theta}$	E1
$\cot 2\theta = 1 + \tan\theta \Rightarrow \dfrac{1-\tan^2\theta}{2\tan\theta} = 1+\tan\theta$
$\Rightarrow 1-\tan^2\theta = 2\tan\theta + 2\tan^2\theta$
$\Rightarrow 3\tan^2\theta + 2\tan\theta - 1 = 0$	M1	Quadratic $= 0$
$\Rightarrow (3\tan\theta - 1)(\tan\theta + 1) = 0$	M1	Factorising or solving
$\tan\theta = \frac{1}{3},\quad \theta = 18.43°, 198.43°$	A3,2,1,0	$18.43°, 198.43°, 135°, 315°$; $-1$ per extra solution in range
or $\tan\theta = -1,\quad \theta = 135°, 315°$
[7]

## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$ | M1 | oe, e.g. converting either side into a one-line fraction(s) involving $\sin\theta$ and $\cos\theta$ |
| $\cot 2\theta = \dfrac{1}{\tan 2\theta} = \dfrac{1-\tan^2\theta}{2\tan\theta}$ | E1 | |
| $\cot 2\theta = 1 + \tan\theta \Rightarrow \dfrac{1-\tan^2\theta}{2\tan\theta} = 1+\tan\theta$ | | |
| $\Rightarrow 1-\tan^2\theta = 2\tan\theta + 2\tan^2\theta$ | | |
| $\Rightarrow 3\tan^2\theta + 2\tan\theta - 1 = 0$ | M1 | Quadratic $= 0$ |
| $\Rightarrow (3\tan\theta - 1)(\tan\theta + 1) = 0$ | M1 | Factorising or solving |
| $\tan\theta = \frac{1}{3},\quad \theta = 18.43°, 198.43°$ | A3,2,1,0 | $18.43°, 198.43°, 135°, 315°$; $-1$ per extra solution in range |
| or $\tan\theta = -1,\quad \theta = 135°, 315°$ | | |
| **[7]** | | |

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4 Show that $\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }$.\\
Hence solve the equation

$$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$

\hfill \mbox{\textit{OCR MEI C4  Q4 [7]}}

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Q1 6 Q2 4 Q3 18 Q4 7 Q5 3