CAIE P1 2008 November — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2008
SessionNovember
Marks4
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Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve algebraic trigonometric identity
DifficultyModerate -0.8 This is a straightforward algebraic manipulation of a trigonometric identity requiring students to find a common denominator and simplify. The technique is standard (multiply fractions, combine, use Pythagorean identity sin²x + cos²x = 1), with no conceptual difficulty or novel insight required. Easier than average as it's a routine P1-level proof with clear algebraic steps.
Spec1.01a Proof: structure of mathematical proof and logical steps1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1
2 Prove the identity $$\frac { 1 + \sin x } { \cos x } + \frac { \cos x } { 1 + \sin x } \equiv \frac { 2 } { \cos x }$$
AnswerMarks Guidance
\(\text{LHS} = \frac{(1+\sin x)^2 + \cos^2 x}{\cos x(1 + \sin x)}\)M1 Reasonable algebra. Correct denominator and one term correct in numerator
\(= \frac{2 + 2\sin x}{\cos x(1 + \sin x)}\)M1, A1 Use of \(\sin^2 x + \cos^2 x = 1\). For \(2 + 2\sin x\)
\(= \frac{2}{\cos x}\)A1 Co – answer was given – check preceding line 
$\text{LHS} = \frac{(1+\sin x)^2 + \cos^2 x}{\cos x(1 + \sin x)}$ | M1 | Reasonable algebra. Correct denominator and one term correct in numerator

$= \frac{2 + 2\sin x}{\cos x(1 + \sin x)}$ | M1, A1 | Use of $\sin^2 x + \cos^2 x = 1$. For $2 + 2\sin x$

$= \frac{2}{\cos x}$ | A1 | Co – answer was given – check preceding line

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2 Prove the identity

$$\frac { 1 + \sin x } { \cos x } + \frac { \cos x } { 1 + \sin x } \equiv \frac { 2 } { \cos x }$$

\hfill \mbox{\textit{CAIE P1 2008 Q2 [4]}}
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Q1 3 Q2 4 Q3 4 Q4 7 Q5 8 Q6 8 Q7 8 Q8 9 Q9 12 Q10 12