| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) requires algebraic manipulation of fractions with a common denominator and applying the Pythagorean identity, which is standard bookwork. Part (b) is a straightforward application using tan θ = ±2, then finding angles in the given range. This is slightly above average due to the two-part structure and fraction manipulation, but remains a routine exercise in trigonometric identities and equation solving. |
| 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 |
|---|---|---|
| \(\left(\frac{\sin\theta}{1-\sin\theta} - \frac{\sin\theta}{1+\sin\theta} =\right) \frac{\sin\theta(1+\sin\theta) - \sin\theta(1-\sin\theta)}{1-\sin^2\theta}\) | *M1 | Put over a single common denominator |
| \(\frac{2\sin^2\theta}{\cos^2\theta}\) | DM1 | Replace \(1-\sin^2\theta\) by \(\cos^2\theta\) and simplify numerator |
| \(2\tan^2\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\tan^2\theta = 8 \rightarrow \tan\theta = (\pm)2\) | B1 | SOI |
| \((\theta =)\ 63.4°,\ 116.6°\) | B1 B1 FT | FT on \(180 -\) 1st solution (with justification) |
## Question 7(a):
| $\left(\frac{\sin\theta}{1-\sin\theta} - \frac{\sin\theta}{1+\sin\theta} =\right) \frac{\sin\theta(1+\sin\theta) - \sin\theta(1-\sin\theta)}{1-\sin^2\theta}$ | *M1 | Put over a single common denominator |
|---|---|---|
| $\frac{2\sin^2\theta}{\cos^2\theta}$ | DM1 | Replace $1-\sin^2\theta$ by $\cos^2\theta$ and simplify numerator |
| $2\tan^2\theta$ | A1 | AG |
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## Question 7(b):
| $2\tan^2\theta = 8 \rightarrow \tan\theta = (\pm)2$ | B1 | SOI |
|---|---|---|
| $(\theta =)\ 63.4°,\ 116.6°$ | B1 B1 FT | FT on $180 -$ 1st solution (with justification) |
---7
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } \equiv 2 \tan ^ { 2 } \theta$.
\item Hence solve the equation $\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } = 8$, for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q7 [6]}}