| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Substitution t equals tan |
| Difficulty | Challenging +1.2 This is a Further Maths FP1 question requiring knowledge of the t = tan(x/2) substitution formulas and algebraic manipulation to prove identities. While it involves multiple steps and Further Maths content (inherently harder), the question is highly structured with part (a) feeding directly into part (b), and the manipulations are relatively straightforward once the standard substitution formulas are recalled. It's above average due to the Further Maths context and multi-step proof nature, but not exceptionally challenging as it follows a clear path with guided steps. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Uses \(\sec x\) and the \(t\)-substitutions for both \(\cos x\) and \(\tan x\) to obtain an expression in terms of \(t\) | 2.1 |
| M1 | Sorts out the \(\sec x\) term, and puts over a common denominator of \(1-t^2\) | 1.1b |
| A1* | Factorises both numerator and denominator (must be seen) and cancels the \((1-t)\) term to achieve the answer | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Uses the \(t\)-substitution for \(\sin x\) in both numerator and denominator | 1.1a |
| M1 | Multiplies through by \(1-t^2\) in numerator and denominator | 1.1b |
| A1* | Factorises both numerator and denominator and makes the connection with part (a) to achieve the given result | 2.1 |
# Question 1
## 1(a)
**M1** | Uses $\sec x$ and the $t$-substitutions for both $\cos x$ and $\tan x$ to obtain an expression in terms of $t$ | 2.1
**M1** | Sorts out the $\sec x$ term, and puts over a common denominator of $1-t^2$ | 1.1b
**A1*** | Factorises both numerator and denominator (must be seen) and cancels the $(1-t)$ term to achieve the answer | 2.1
(3 marks)
## 1(b)
**M1** | Uses the $t$-substitution for $\sin x$ in both numerator and denominator | 1.1a
**M1** | Multiplies through by $1-t^2$ in numerator and denominator | 1.1b
**A1*** | Factorises both numerator and denominator and makes the connection with part (a) to achieve the given result | 2.1
(3 marks)
**Total: 6 marks**\begin{enumerate}
\item (a) Use the substitution $\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)$ to show that
\end{enumerate}
$$\sec x - \tan x \equiv \frac { 1 - t } { 1 + t } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$
(b) Use the substitution $\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)$ and the answer to part (a) to prove that
$$\frac { 1 - \sin x } { 1 + \sin x } \equiv ( \sec x - \tan x ) ^ { 2 } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$
\section*{Q uestion 1 continued}
\hfill \mbox{\textit{Edexcel FP1 AS Q1 [6]}}