| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Prove identity then solve |
| Difficulty | Standard +0.3 Part (a) is a standard algebraic manipulation of trigonometric identities requiring common denominator work and basic identities (tan θ = sin θ/cos θ). Part (b) uses the proven identity to create a solvable equation, requiring substitution and solving a quadratic in sin θ. This is routine bookwork with straightforward algebraic steps, slightly easier than average due to the scaffolding provided by part (a). |
| 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 |
|---|---|---|
| 7(a): \(\frac{\tan\theta}{1+\cos\theta} + \frac{\tan\theta}{1-\cos\theta} = \frac{\tan\theta(1-\cos\theta) + \tan\theta(1+\cos\theta)}{1-\cos^2\theta}\) | M1 | |
| \(= \frac{2\tan\theta}{\sin^2\theta}\) | M1 | |
| \(= \frac{2\sin\theta}{\cos\theta\sin^2\theta}\) | M1 | |
| \(= \frac{2}{\sin\theta\cos\theta}\) | A1 | AG |
| 7(b): \(\frac{2}{\sin\theta\cos\theta} = \frac{6\cos\theta}{\sin\theta}\) | M1 | |
| \(\cos^2\theta = \frac{1}{3} \rightarrow \cos\theta = (\pm)0.5774\) | A1 | |
| \(54.7°,\ 125.3°\) (FT for \(180° -\) 1st solution) | A1, A1FT |
## Question 7:
**7(a):** $\frac{\tan\theta}{1+\cos\theta} + \frac{\tan\theta}{1-\cos\theta} = \frac{\tan\theta(1-\cos\theta) + \tan\theta(1+\cos\theta)}{1-\cos^2\theta}$ | M1 |
$= \frac{2\tan\theta}{\sin^2\theta}$ | M1 |
$= \frac{2\sin\theta}{\cos\theta\sin^2\theta}$ | M1 |
$= \frac{2}{\sin\theta\cos\theta}$ | A1 | AG
**7(b):** $\frac{2}{\sin\theta\cos\theta} = \frac{6\cos\theta}{\sin\theta}$ | M1 |
$\cos^2\theta = \frac{1}{3} \rightarrow \cos\theta = (\pm)0.5774$ | A1 |
$54.7°,\ 125.3°$ (FT for $180° -$ 1st solution) | A1, A1FT |
---7
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } \equiv \frac { 2 } { \sin \theta \cos \theta }$.
\item Hence solve the equation $\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } = \frac { 6 } { \tan \theta }$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q7 [8]}}