Question 1 - A-Level Maths

Edexcel C3 2005 June — Question 1 8 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2005
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Solve equation using Pythagorean identities
Difficulty	Moderate -0.3 Part (a) is a standard bookwork proof requiring simple division by cos²θ. Part (b) is a routine application of the proven identity to convert to a quadratic in secθ, then solve. This is a typical C3 exercise testing standard technique with no novel insight required, making it slightly easier than average.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 1.05o Trigonometric equations: solve in given intervals

(a) Given that $\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1$, show that $1 + \tan ^ { 2 } \theta \equiv \sec ^ { 2 } \theta$.
(b) Solve, for $0 \leqslant \theta < 360 ^ { \circ }$, the equation

$$2 \tan ^ { 2 } \theta + \sec \theta = 1 ,$$ giving your answers to 1 decimal place.

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Question 1:

Part (a):

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
Dividing by $\cos^2\theta$: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}$	M1
Completion: $1 + \tan^2\theta \equiv \sec^2\theta$	A1	No errors seen

Part (b):

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
Use of $1 + \tan^2\theta = \sec^2\theta$: $2(\sec^2\theta - 1) + \sec\theta = 1$, giving $[2\sec^2\theta + \sec\theta - 3 = 0]$	M1
Factorising or solving: $(2\sec\theta + 3)(\sec\theta - 1) = 0$, $[\sec\theta = -\frac{3}{2}$ or $\sec\theta = 1]$	M1
$\theta = 0$	B1
$\cos\theta = -\frac{2}{3}$; $\theta_1 = 131.8°$	M1 A1
$\theta_2 = 228.2°$	A1ft	Follow through for $\theta_2 = 360° - \theta_1$

## Question 1:

### Part (a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Dividing by $\cos^2\theta$: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}$ | M1 | |
| Completion: $1 + \tan^2\theta \equiv \sec^2\theta$ | A1 | No errors seen |

### Part (b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Use of $1 + \tan^2\theta = \sec^2\theta$: $2(\sec^2\theta - 1) + \sec\theta = 1$, giving $[2\sec^2\theta + \sec\theta - 3 = 0]$ | M1 | |
| Factorising or solving: $(2\sec\theta + 3)(\sec\theta - 1) = 0$, $[\sec\theta = -\frac{3}{2}$ or $\sec\theta = 1]$ | M1 | |
| $\theta = 0$ | B1 | |
| $\cos\theta = -\frac{2}{3}$; $\theta_1 = 131.8°$ | M1 A1 | |
| $\theta_2 = 228.2°$ | A1ft | Follow through for $\theta_2 = 360° - \theta_1$ |

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Show LaTeX source

\begin{enumerate}
  \item (a) Given that $\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1$, show that $1 + \tan ^ { 2 } \theta \equiv \sec ^ { 2 } \theta$.\\
(b) Solve, for $0 \leqslant \theta < 360 ^ { \circ }$, the equation
\end{enumerate}

$$2 \tan ^ { 2 } \theta + \sec \theta = 1 ,$$

giving your answers to 1 decimal place.\\

\hfill \mbox{\textit{Edexcel C3 2005 Q1 [8]}}

This paper (7 questions)

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Q1 8 Q2 12 Q3 10 Q4 9 Q5 15 Q6 11 Q7 10

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
Use of \(1 + \tan^2\theta = \sec^2\theta\): \(2(\sec^2\theta - 1) + \sec\theta = 1\), giving \([2\sec^2\theta + \sec\theta - 3 = 0]\)	M1
Factorising or solving: \((2\sec\theta + 3)(\sec\theta - 1) = 0\), \([\sec\theta = -\frac{3}{2}\) or \(\sec\theta = 1]\)	M1
\(\theta = 0\)	B1
\(\cos\theta = -\frac{2}{3}\); \(\theta_1 = 131.8°\)	M1 A1
\(\theta_2 = 228.2°\)	A1ft	Follow through for \(\theta_2 = 360° - \theta_1\)