CAIE P1 2007 June — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2007
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve algebraic trigonometric identity
DifficultyModerate -0.3 This is a straightforward identity proof requiring standard techniques: recognizing 1 + tan²x ≡ sec²x, converting tan to sin/cos, and simplifying using cos²x + sin²x = 1. It's slightly easier than average as it follows a predictable pattern with no novel insight required, though it does require competent algebraic manipulation across multiple steps.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2
3 Prove the identity \(\frac { 1 - \tan ^ { 2 } x } { 1 + \tan ^ { 2 } x } \equiv 1 - 2 \sin ^ { 2 } x\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Use of \(t = s/c\)M1 \(\tan\) completely removed
\(\rightarrow (c^2 - s^2) \div (c^2 + s^2)\)A1 May omit the denominator (\(=1\))
Use of \(c^2 + s^2 = 1\)M1 Whenever used appropriately
\(\rightarrow (c^2 - s^2) \rightarrow 1 - 2\sin^2 x\)A1 [4] AG – Beware fortuitous answers 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $t = s/c$ | M1 | $\tan$ completely removed |
| $\rightarrow (c^2 - s^2) \div (c^2 + s^2)$ | A1 | May omit the denominator ($=1$) |
| Use of $c^2 + s^2 = 1$ | M1 | Whenever used appropriately |
| $\rightarrow (c^2 - s^2) \rightarrow 1 - 2\sin^2 x$ | A1 [4] | AG – Beware fortuitous answers |

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3 Prove the identity $\frac { 1 - \tan ^ { 2 } x } { 1 + \tan ^ { 2 } x } \equiv 1 - 2 \sin ^ { 2 } x$.

\hfill \mbox{\textit{CAIE P1 2007 Q3 [4]}}
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Q1 4 Q2 4 Q3 4 Q4 4 Q5 5 Q6 7 Q7 7 Q8 8 Q9 8 Q10 12 Q11 12