CAIE P1 2009 June — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2009
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve algebraic trigonometric identity
DifficultyStandard +0.3 This is a straightforward algebraic manipulation of a trigonometric identity requiring students to find a common denominator, simplify the numerator, and apply the Pythagorean identity sin²x + cos²x = 1 to reach tan²x. While it involves multiple steps, the techniques are standard and the path is relatively clear, making it slightly easier than average.
Spec1.05p Proof involving trig: functions and identities
1 Prove the identity \(\frac { \sin x } { 1 - \sin x } - \frac { \sin x } { 1 + \sin x } \equiv 2 \tan ^ { 2 } x\).
AnswerMarks Guidance
\(\frac{s}{1-s} + \frac{s}{1+s} = \frac{2x^2}{1-s^2}\)B1 Correct algebra
Use of \(1 - s^2 = c^2\)M1 Use of this formula
\(\rightarrow \frac{2x^2}{c^2}\)A1 Evidence of \(\tan = \sin/\cos\) and everything completed accurately
\(\rightarrow 2t^2\)[3] 
$\frac{s}{1-s} + \frac{s}{1+s} = \frac{2x^2}{1-s^2}$ | B1 | Correct algebra
Use of $1 - s^2 = c^2$ | M1 | Use of this formula
$\rightarrow \frac{2x^2}{c^2}$ | A1 | Evidence of $\tan = \sin/\cos$ and everything completed accurately
$\rightarrow 2t^2$ | [3] |
1 Prove the identity $\frac { \sin x } { 1 - \sin x } - \frac { \sin x } { 1 + \sin x } \equiv 2 \tan ^ { 2 } x$.

\hfill \mbox{\textit{CAIE P1 2009 Q1 [3]}}
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