Question 2 - A-Level Maths

CAIE P1 2010 November — Question 2 4 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2010
Session	November
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Prove algebraic trigonometric identity
Difficulty	Moderate -0.5 This is a straightforward algebraic manipulation of a trigonometric identity requiring students to express tan²x as sin²x/cos²x, factor out sin²x, and use the Pythagorean identity. It's slightly easier than average because it's a direct proof with a clear path: no problem-solving insight needed, just systematic application of standard identities.
Spec	1.01a Proof: structure of mathematical proof and logical steps 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

2 Prove the identity $$\tan ^ { 2 } x - \sin ^ { 2 } x \equiv \tan ^ { 2 } x \sin ^ { 2 } x$$

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Question 2:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\text{LHS} = \sin^2x/\cos^2x - \sin^2x$	M1	Replace $t^2$ by $s^2/c^2$ or $\sec^2 - 1$
$\sin^2x(1-\cos^2x)/\cos^2x$	M1	Use of $1 - \cos^2x = \sin^2x$
$\dfrac{\sin^2x \sin^2x}{\cos^2x}$ oe	M1	Valid overall method
$\tan^2x\sin^2x$	A1	AG
OR RHS $= \dfrac{\sin^2x}{\cos^2x} \cdot \sin^2x$	M1	Replace $t^2$ by $s^2/c^2$
$\sin^2x(1-\cos^2x)/\cos^2x$	M1	Use of $1-\cos^2x = \sin^2x$
$(\sin^2x/\cos^2x) - \sin^2x$	M1	Valid overall method
$\tan^2x - \sin^2x$	A1	AG

## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{LHS} = \sin^2x/\cos^2x - \sin^2x$ | M1 | Replace $t^2$ by $s^2/c^2$ or $\sec^2 - 1$ |
| $\sin^2x(1-\cos^2x)/\cos^2x$ | M1 | Use of $1 - \cos^2x = \sin^2x$ |
| $\dfrac{\sin^2x \sin^2x}{\cos^2x}$ oe | M1 | Valid overall method |
| $\tan^2x\sin^2x$ | A1 | **AG** |
| **OR RHS** $= \dfrac{\sin^2x}{\cos^2x} \cdot \sin^2x$ | M1 | Replace $t^2$ by $s^2/c^2$ |
| $\sin^2x(1-\cos^2x)/\cos^2x$ | M1 | Use of $1-\cos^2x = \sin^2x$ |
| $(\sin^2x/\cos^2x) - \sin^2x$ | M1 | Valid overall method |
| $\tan^2x - \sin^2x$ | A1 | **AG** |

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2 Prove the identity

$$\tan ^ { 2 } x - \sin ^ { 2 } x \equiv \tan ^ { 2 } x \sin ^ { 2 } x$$

\hfill \mbox{\textit{CAIE P1 2010 Q2 [4]}}

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Q1 4 Q2 4 Q3 4 Q4 5 Q5 6 Q6 7 Q7 7 Q8 7 Q9 8 Q10 10 Q11 13

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(\text{LHS} = \sin^2x/\cos^2x - \sin^2x\)	M1	Replace \(t^2\) by \(s^2/c^2\) or \(\sec^2 - 1\)
\(\sin^2x(1-\cos^2x)/\cos^2x\)	M1	Use of \(1 - \cos^2x = \sin^2x\)
\(\dfrac{\sin^2x \sin^2x}{\cos^2x}\) oe	M1	Valid overall method
\(\tan^2x\sin^2x\)	A1	AG
OR RHS \(= \dfrac{\sin^2x}{\cos^2x} \cdot \sin^2x\)	M1	Replace \(t^2\) by \(s^2/c^2\)
\(\sin^2x(1-\cos^2x)/\cos^2x\)	M1	Use of \(1-\cos^2x = \sin^2x\)
\((\sin^2x/\cos^2x) - \sin^2x\)	M1	Valid overall method
\(\tan^2x - \sin^2x\)	A1	AG