CAIE P2 2018 November — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2018
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeSolve equation using Pythagorean identities
DifficultyStandard +0.3 This question requires converting sec²θ to 1/cos²θ and cosec θ to 1/sin θ, then using the Pythagorean identity to obtain a quadratic in sin θ. While it involves multiple steps and reciprocal trig functions, it follows a standard pattern for this topic with straightforward algebraic manipulation. The 5-mark allocation and restricted domain make it slightly easier than average.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals
Solve the equation \(\sec^2 \theta = 3 \cosec \theta\) for \(0° < \theta < 180°\). [5]
Question 3:
AnswerMarks
31 3 3
State = or 1+tan2θ=
AnswerMarks Guidance
cos2θ sinθ sinθB1
Produce quadratic equation in sinθM1 Dependent on B1
Solve 3-term quadratic equation to find value between –1 and 1 for sinθM1 Dependent on first M1
Obtain sinθ= 1(−1+ 37) and hence 57.9
AnswerMarks
6A1
Obtain 122.1 and no others between 0 and 180A1
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
3 | 1 3 3
State = or 1+tan2θ=
cos2θ sinθ sinθ | B1
Produce quadratic equation in sinθ | M1 | Dependent on B1
Solve 3-term quadratic equation to find value between –1 and 1 for sinθ | M1 | Dependent on first M1
Obtain sinθ= 1(−1+ 37) and hence 57.9
6 | A1
Obtain 122.1 and no others between 0 and 180 | A1
5
Question | Answer | Marks | Guidance
Solve the equation $\sec^2 \theta = 3 \cosec \theta$ for $0° < \theta < 180°$. [5]

\hfill \mbox{\textit{CAIE P2 2018 Q3 [5]}}
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Q1 5 Q2 5 Q3 5 Q4 11 Q5 9 Q6 8 Q7 10