| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear signposting ('show that', 'hence'). Part (a) requires algebraic manipulation using tan x = sin x/cos x and factoring, which is straightforward. Parts (b) and (c) involve routine rearrangement and solving a simple equation. While it requires multiple steps, each step follows standard techniques without requiring novel insight, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\tan x + \sin x}{\tan x - \sin x}[=k]\) leading to \(\frac{\sin x + \sin x \cos x}{\sin x - \sin x \cos x}[=k]\) or \(\dfrac{\frac{1}{\cos x}+1}{\frac{1}{\cos x}-1}[=k]\) or \(\frac{\tan x + \tan x \cos x}{\tan x - \tan x \cos x}[=k]\) | M1 | Multiply numerator and denominator by \(\cos x\), or divide numerator and denominator by \(\tan x\) or \(\sin x\) |
| \(\frac{\sin x(1+\cos x)}{\sin x(1-\cos x)}\) or \(\dfrac{\frac{1}{\cos x}+1}{\frac{1}{\cos x}-1} \cdot \frac{\cos x}{\cos x}\) or \(\frac{\tan x(1+\cos x)}{\tan x(1-\cos x)}\) leading to \(\frac{1+\cos x}{1-\cos x}[=k]\) | A1 | AG, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k - k\cos x = 1 + \cos x\) leading to \(k - 1 = k\cos x + \cos x\) | M1 | Gather like terms on LHS and RHS |
| \(k - 1 = (k+1)\cos x\) leading to \(\cos x = \dfrac{k-1}{k+1}\) | A1 | WWW, OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtaining \(\cos x\) from their (b) or (a) | M1 | Expect \(\cos x = \dfrac{3}{5}\) |
| \(\pm 0.927\) (only solutions in the given range) | A1 | AWRT. Accept \(\pm 0.295\pi\) |
## Question 4:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\tan x + \sin x}{\tan x - \sin x}[=k]$ leading to $\frac{\sin x + \sin x \cos x}{\sin x - \sin x \cos x}[=k]$ or $\dfrac{\frac{1}{\cos x}+1}{\frac{1}{\cos x}-1}[=k]$ or $\frac{\tan x + \tan x \cos x}{\tan x - \tan x \cos x}[=k]$ | M1 | Multiply numerator and denominator by $\cos x$, or divide numerator and denominator by $\tan x$ or $\sin x$ |
| $\frac{\sin x(1+\cos x)}{\sin x(1-\cos x)}$ or $\dfrac{\frac{1}{\cos x}+1}{\frac{1}{\cos x}-1} \cdot \frac{\cos x}{\cos x}$ or $\frac{\tan x(1+\cos x)}{\tan x(1-\cos x)}$ leading to $\frac{1+\cos x}{1-\cos x}[=k]$ | A1 | AG, WWW |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k - k\cos x = 1 + \cos x$ leading to $k - 1 = k\cos x + \cos x$ | M1 | Gather like terms on LHS and RHS |
| $k - 1 = (k+1)\cos x$ leading to $\cos x = \dfrac{k-1}{k+1}$ | A1 | WWW, OE |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtaining $\cos x$ from their **(b)** or **(a)** | M1 | Expect $\cos x = \dfrac{3}{5}$ |
| $\pm 0.927$ (only solutions in the given range) | A1 | AWRT. Accept $\pm 0.295\pi$ |
---4
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$\frac { \tan x + \sin x } { \tan x - \sin x } = k$$
where $k$ is a constant, may be expressed as
$$\frac { 1 + \cos x } { 1 - \cos x } = k$$
\item Hence express $\cos x$ in terms of $k$.
\item Hence solve the equation $\frac { \tan x + \sin x } { \tan x - \sin x } = 4$ for $- \pi < x < \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q4 [6]}}