Question 5 - A-Level Maths

CAIE P1 2020 March — Question 5 5 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2020
Session	March
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Standard trigonometric equations
Type	Rational trig expressions
Difficulty	Standard +0.8 This requires algebraic manipulation of a rational trig expression, converting tan to sin/cos, clearing denominators, and solving a resulting trigonometric equation. The multi-step algebraic work and need to handle the tan θ = sin θ/cos θ substitution carefully makes this moderately challenging, though the restricted domain simplifies the solution process.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

5 Solve the equation $$\frac { \tan \theta + 3 \sin \theta + 2 } { \tan \theta - 3 \sin \theta + 1 } = 2$$ for $0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }$.

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Question 5:

Answer	Marks	Guidance
$2\tan\theta - 6\sin\theta + 2 = \tan\theta + 3\sin\theta + 2 \to \tan\theta - 9\sin\theta\,(=0)$	M1	Multiply by denominator and simplify
$\sin\theta - 9\sin\theta\cos\theta\,(=0)$	M1	Multiply by $\cos\theta$
$\sin\theta(1-9\cos\theta)\,(=0) \to \sin\theta=0,\quad \cos\theta=\frac{1}{9}$	M1	Factorise and attempt to solve at least one of the factors $= 0$
$\theta = 0$ or $83.6°$ (only answers in the given range)	A1A1
Total: 5

## Question 5:

| $2\tan\theta - 6\sin\theta + 2 = \tan\theta + 3\sin\theta + 2 \to \tan\theta - 9\sin\theta\,(=0)$ | M1 | Multiply by denominator and simplify |
|---|---|---|
| $\sin\theta - 9\sin\theta\cos\theta\,(=0)$ | M1 | Multiply by $\cos\theta$ |
| $\sin\theta(1-9\cos\theta)\,(=0) \to \sin\theta=0,\quad \cos\theta=\frac{1}{9}$ | M1 | Factorise and attempt to solve at least one of the factors $= 0$ |
| $\theta = 0$ or $83.6°$ (only answers in the given range) | A1A1 | |
| **Total: 5** | | |

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5 Solve the equation

$$\frac { \tan \theta + 3 \sin \theta + 2 } { \tan \theta - 3 \sin \theta + 1 } = 2$$

for $0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P1 2020 Q5 [5]}}

This paper (12 questions)

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Q1 3 Q2 4 Q3 4 Q4 4 Q5 5 Q6 5 Q7 6 Q8 6 Q9 8 Q10 10 Q11 9 Q12 11

Answer	Marks	Guidance
\(2\tan\theta - 6\sin\theta + 2 = \tan\theta + 3\sin\theta + 2 \to \tan\theta - 9\sin\theta\,(=0)\)	M1	Multiply by denominator and simplify
\(\sin\theta - 9\sin\theta\cos\theta\,(=0)\)	M1	Multiply by \(\cos\theta\)
\(\sin\theta(1-9\cos\theta)\,(=0) \to \sin\theta=0,\quad \cos\theta=\frac{1}{9}\)	M1	Factorise and attempt to solve at least one of the factors \(= 0\)
\(\theta = 0\) or \(83.6°\) (only answers in the given range)	A1A1
Total: 5