| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a standard Further Maths Pure identity proof followed by a routine substitution. Part (a) requires algebraic manipulation with reciprocal trig functions (combining fractions, using Pythagorean identities), which is straightforward for FM students. Part (b) is a direct application using the 'hence' structure, requiring only substitution of 2x and solving a simple equation. The techniques are well-practiced and the question follows a predictable pattern, making it slightly easier than average even for Further Maths. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{\cosec\theta-1}+\frac{1}{\cosec\theta+1}\right) \equiv \frac{\cosec\theta+1+\cosec\theta-1}{(\cosec\theta-1)(\cosec\theta+1)}\) | B1 | Adds fractions to obtain a correct single fraction (not fractions over fractions). Condone missing brackets if recovered. |
| \(\equiv \frac{2\cosec\theta}{\cosec^2\theta-1} \equiv \frac{2\cosec\theta}{\cot^2\theta}\) or e.g. \(\equiv \frac{2\sin\theta}{1-\sin^2\theta} \equiv \frac{2\sin\theta}{\cos^2\theta}\) | M1 | Uses a correct Pythagorean identity e.g. \(\cosec^2\theta-1=\cot^2\theta\), \(\sin^2\theta+\cos^2\theta=1\) |
| \(\frac{2\cosec\theta}{\cot^2\theta} \equiv \frac{2}{\sin\theta}\times\frac{\sin^2\theta}{\cos^2\theta} \equiv 2\tan\theta\sec\theta\)* or \(\frac{2\sin\theta}{\cos^2\theta} \equiv 2\tan\theta\sec\theta\)* | A1* | Correct work with all necessary steps shown. No notational/bracketing errors, no mixed/missing variables. Condone reaching \(2\sec\theta\tan\theta\)* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{LHS} = \frac{1}{\cosec\theta-1}+\frac{1}{\cosec\theta+1} \equiv \frac{\cosec\theta+1+\cosec\theta-1}{(\cosec\theta-1)(\cosec\theta+1)}\) | B1 | See Way 1 |
| \(\equiv \frac{2\cosec\theta}{\cosec^2\theta-1} \equiv \frac{2\cosec\theta}{\cot^2\theta}\) | M1 | See Way 1 |
| \(\text{RHS} = 2\tan\theta\sec\theta \equiv \frac{2\sin\theta}{\cos^2\theta} \equiv \frac{2\sin^2\theta}{\sin\theta\cos^2\theta} \equiv \frac{2}{\sin\theta}\times\frac{\sin^2\theta}{\cos^2\theta} \equiv \frac{2\cosec\theta}{\cot^2\theta} = \text{LHS}\) | A1* | Must have a minimal conclusion e.g. "= LHS", "QED", "Hence proven". No notational/bracketing errors. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\tan 2x\sec 2x = \cot 2x\sec 2x\) (equation deduced from part (a)) | B1 | Deduces the correct equation using the result from part (a) |
| \(\sec 2x(2\tan 2x - \cot 2x)=0\), leading to e.g. \(\tan^2 2x = \frac{1}{2}\), or \(\sin^2 2x = \frac{1}{3}\), or \(\cos^2 2x = \frac{2}{3}\), or \(\cos 4x = \frac{1}{3}\), or \(\tan^4 x - 10\tan^2 x +1=0\) | M1 | Factors/cancels \(\sec 2x\) to obtain equation of form \(\tan^2 2x = \alpha\), \(\sin^2 2x = \beta\), \(\cos^2 2x = \gamma\), or \(\cos 4x = k\) where \(\alpha>0\), \(0<\beta<1\), \(0<\gamma<1\), \(0 |
| \(2x = \tan^{-1}\frac{1}{\sqrt{2}} = K \Rightarrow x = \frac{K}{2}\), or \(2x=\sin^{-1}\frac{1}{\sqrt{3}}=K \Rightarrow x=\frac{K}{2}\), or \(2x=\cos^{-1}\sqrt{\frac{2}{3}}=K \Rightarrow x=\frac{K}{2}\), or \(4x=\cos^{-1}\left(\frac{1}{3}\right)=K \Rightarrow x=\frac{1}{4}K\) | M1 | Correct order of operations leading to at least one value for \(x\). May be implied by 17.6° or 17.7° with no incorrect work. |
| \(x = 17.6°,\ 72.4°\) | A1 | Allow awrt \(17.6°\) and awrt \(72.4°\). Ignore values outside range; extra angles in range score A0. Answers in radians score A0. |
## Question 8(a):
**Way 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{\cosec\theta-1}+\frac{1}{\cosec\theta+1}\right) \equiv \frac{\cosec\theta+1+\cosec\theta-1}{(\cosec\theta-1)(\cosec\theta+1)}$ | B1 | Adds fractions to obtain a correct single fraction (not fractions over fractions). Condone missing brackets if recovered. |
| $\equiv \frac{2\cosec\theta}{\cosec^2\theta-1} \equiv \frac{2\cosec\theta}{\cot^2\theta}$ or e.g. $\equiv \frac{2\sin\theta}{1-\sin^2\theta} \equiv \frac{2\sin\theta}{\cos^2\theta}$ | M1 | Uses a correct Pythagorean identity e.g. $\cosec^2\theta-1=\cot^2\theta$, $\sin^2\theta+\cos^2\theta=1$ |
| $\frac{2\cosec\theta}{\cot^2\theta} \equiv \frac{2}{\sin\theta}\times\frac{\sin^2\theta}{\cos^2\theta} \equiv 2\tan\theta\sec\theta$* or $\frac{2\sin\theta}{\cos^2\theta} \equiv 2\tan\theta\sec\theta$* | A1* | Correct work with all necessary steps shown. No notational/bracketing errors, no mixed/missing variables. Condone reaching $2\sec\theta\tan\theta$* |
**Way 2 ("Meets in the middle"):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{LHS} = \frac{1}{\cosec\theta-1}+\frac{1}{\cosec\theta+1} \equiv \frac{\cosec\theta+1+\cosec\theta-1}{(\cosec\theta-1)(\cosec\theta+1)}$ | B1 | See Way 1 |
| $\equiv \frac{2\cosec\theta}{\cosec^2\theta-1} \equiv \frac{2\cosec\theta}{\cot^2\theta}$ | M1 | See Way 1 |
| $\text{RHS} = 2\tan\theta\sec\theta \equiv \frac{2\sin\theta}{\cos^2\theta} \equiv \frac{2\sin^2\theta}{\sin\theta\cos^2\theta} \equiv \frac{2}{\sin\theta}\times\frac{\sin^2\theta}{\cos^2\theta} \equiv \frac{2\cosec\theta}{\cot^2\theta} = \text{LHS}$ | A1* | Must have a minimal conclusion e.g. "= LHS", "QED", "Hence proven". No notational/bracketing errors. |
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## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\tan 2x\sec 2x = \cot 2x\sec 2x$ (equation deduced from part (a)) | B1 | Deduces the correct equation using the result from part (a) |
| $\sec 2x(2\tan 2x - \cot 2x)=0$, leading to e.g. $\tan^2 2x = \frac{1}{2}$, or $\sin^2 2x = \frac{1}{3}$, or $\cos^2 2x = \frac{2}{3}$, or $\cos 4x = \frac{1}{3}$, or $\tan^4 x - 10\tan^2 x +1=0$ | M1 | Factors/cancels $\sec 2x$ to obtain equation of form $\tan^2 2x = \alpha$, $\sin^2 2x = \beta$, $\cos^2 2x = \gamma$, or $\cos 4x = k$ where $\alpha>0$, $0<\beta<1$, $0<\gamma<1$, $0<k<1$ |
| $2x = \tan^{-1}\frac{1}{\sqrt{2}} = K \Rightarrow x = \frac{K}{2}$, or $2x=\sin^{-1}\frac{1}{\sqrt{3}}=K \Rightarrow x=\frac{K}{2}$, or $2x=\cos^{-1}\sqrt{\frac{2}{3}}=K \Rightarrow x=\frac{K}{2}$, or $4x=\cos^{-1}\left(\frac{1}{3}\right)=K \Rightarrow x=\frac{1}{4}K$ | M1 | Correct order of operations leading to at least one value for $x$. May be implied by 17.6° or 17.7° with no incorrect work. |
| $x = 17.6°,\ 72.4°$ | A1 | Allow awrt $17.6°$ and awrt $72.4°$. Ignore values outside range; extra angles in range score A0. Answers in radians score A0. |\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Prove that
$$\frac { 1 } { \operatorname { cosec } \theta - 1 } + \frac { 1 } { \operatorname { cosec } \theta + 1 } \equiv 2 \tan \theta \sec \theta \quad \theta \neq ( 90 n ) ^ { \circ } , n \in \mathbb { Z }$$
(b) Hence solve, for $0 < x < 90 ^ { \circ }$, the equation
$$\frac { 1 } { \operatorname { cosec } 2 x - 1 } + \frac { 1 } { \operatorname { cosec } 2 x + 1 } = \cot 2 x \sec 2 x$$
Give each answer, in degrees, to one decimal place.
\hfill \mbox{\textit{Edexcel Paper 2 2024 Q8 [7]}}