Question 5 - A-Level Maths

CAIE P3 2022 March — Question 5 6 marks

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2022
Session	March
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Addition & Double Angle Formulae
Type	Two angles with tan relationships
Difficulty	Standard +0.8 This question requires applying the tan addition formula, substituting the relationship tan α = 3tan β, solving the resulting quadratic equation, then finding two angle pairs that satisfy both conditions. It demands multiple techniques (addition formulae, algebraic manipulation, inverse trig with quadrant consideration) and careful reasoning about which solutions are valid, making it moderately challenging but still within standard P3 scope.
Spec	1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals

5 The angles $\alpha$ and $\beta$ lie between $0 ^ { \circ }$ and $180 ^ { \circ }$ and are such that $$\tan ( \alpha + \beta ) = 2 \quad \text { and } \quad \tan \alpha = 3 \tan \beta .$$ Find the possible values of $\alpha$ and $\beta$.

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Question 5:

Answer	Marks	Guidance
Answer	Mark	Guidance
Use correct $\tan(A+B)$ formula and obtain an equation in $\tan\alpha$ and $\tan\beta$	M1	$\dfrac{\tan + \tan}{1 - \tan\tan} = 2$
Substitute throughout for $\tan\alpha$ or for $\tan\beta$	M1	$\dfrac{3\tan + \tan}{1 - 3\tan^2} = 2$
Obtain $3\tan^2\beta + 2\tan\beta - 1 = 0$ or $\tan^2\alpha + 2\tan\alpha - 3 = 0$	A1	OE e.g. $6\tan^2\beta + 4\tan\beta - 2 = 0$ or $\frac{2}{3}\tan^2\alpha + \frac{4}{3}\tan\alpha - 2 = 0$
Solve a 3-term quadratic and find an angle	M1
Obtain answer $\alpha = 45°$, $\beta = 18.4°$	A1	$\frac{\pi}{4}$ or $0.785$, $0.322$
Obtain answer $\alpha = 108.4°$, $\beta = 135°$	A1	$1.89$, $\frac{3\pi}{4}$ or $2.36$. Answer in radians, max. A1A0 or vice versa. Ignore answers outside $[0°, 180°]$
Total: 6		SC: If A0A0 allow SC B1 for both $\alpha$'s or both $\beta$'s

## Question 5:

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct $\tan(A+B)$ formula and obtain an equation in $\tan\alpha$ and $\tan\beta$ | M1 | $\dfrac{\tan + \tan}{1 - \tan\tan} = 2$ |
| Substitute throughout for $\tan\alpha$ or for $\tan\beta$ | M1 | $\dfrac{3\tan + \tan}{1 - 3\tan^2} = 2$ |
| Obtain $3\tan^2\beta + 2\tan\beta - 1 = 0$ or $\tan^2\alpha + 2\tan\alpha - 3 = 0$ | A1 | OE e.g. $6\tan^2\beta + 4\tan\beta - 2 = 0$ or $\frac{2}{3}\tan^2\alpha + \frac{4}{3}\tan\alpha - 2 = 0$ |
| Solve a 3-term quadratic and find an angle | M1 | |
| Obtain answer $\alpha = 45°$, $\beta = 18.4°$ | A1 | $\frac{\pi}{4}$ or $0.785$, $0.322$ |
| Obtain answer $\alpha = 108.4°$, $\beta = 135°$ | A1 | $1.89$, $\frac{3\pi}{4}$ or $2.36$. Answer in radians, max. A1A0 or vice versa. Ignore answers outside $[0°, 180°]$ |
| **Total: 6** | | SC: If A0A0 allow **SC B1** for both $\alpha$'s or both $\beta$'s |

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5 The angles $\alpha$ and $\beta$ lie between $0 ^ { \circ }$ and $180 ^ { \circ }$ and are such that

$$\tan ( \alpha + \beta ) = 2 \quad \text { and } \quad \tan \alpha = 3 \tan \beta .$$

Find the possible values of $\alpha$ and $\beta$.\\

\hfill \mbox{\textit{CAIE P3 2022 Q5 [6]}}

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