| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) requires straightforward application of Pythagorean identity (sin²θ + cos²θ = 1) and tan²θ = sin²θ/cos²θ to prove the identity through algebraic manipulation. Part (b) uses the proven identity to reduce to a quadratic in tan²θ, then solving within a restricted domain. This is a standard two-part question with routine techniques and no novel insight required, 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 |
| Reach \(\dfrac{\cos^2\theta - \sin^2\theta}{\cos^2\theta}\) or \(\dfrac{1-\sin^2\theta}{1-\sin^2\theta} - \dfrac{\sin^2\theta}{\cos^2\theta}\) or \(\dfrac{\sin^2\theta+\cos^2\theta}{\cos^2\theta} - 2\tan^2\theta\) or \(\sec^2\theta - \dfrac{2\sin^2\theta}{\cos^2\theta}\) or \(2 - \sec^2\theta\) or \(\dfrac{\cos 2\theta}{\cos^2\theta}\) | M1 | May start with \(1 - \tan^2\theta\) |
| \(1 - \tan^2\theta\) | A1 | AG, must show sufficient stages |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - \tan^2\theta = 2\tan^4\theta \Rightarrow 2\tan^4\theta + \tan^2\theta - 1 [=0]\) | M1 | Forming a 3-term quadratic in \(\tan^2\theta\) or e.g. \(u\) |
| \(\tan^2\theta = 0.5\) or \(-1\) leading to \(\tan\theta = [\pm]\sqrt{0.5}\) | M1 | |
| \(\theta = 35.3°\) and \(144.7°\) (AWRT) | A1 | Both correct. Radians \(0.615\), \(2.53\) scores A0. |
| 3 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Reach $\dfrac{\cos^2\theta - \sin^2\theta}{\cos^2\theta}$ or $\dfrac{1-\sin^2\theta}{1-\sin^2\theta} - \dfrac{\sin^2\theta}{\cos^2\theta}$ or $\dfrac{\sin^2\theta+\cos^2\theta}{\cos^2\theta} - 2\tan^2\theta$ or $\sec^2\theta - \dfrac{2\sin^2\theta}{\cos^2\theta}$ or $2 - \sec^2\theta$ or $\dfrac{\cos 2\theta}{\cos^2\theta}$ | M1 | May start with $1 - \tan^2\theta$ |
| $1 - \tan^2\theta$ | A1 | AG, must show sufficient stages |
| | **2** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \tan^2\theta = 2\tan^4\theta \Rightarrow 2\tan^4\theta + \tan^2\theta - 1 [=0]$ | M1 | Forming a 3-term quadratic in $\tan^2\theta$ or e.g. $u$ |
| $\tan^2\theta = 0.5$ or $-1$ leading to $\tan\theta = [\pm]\sqrt{0.5}$ | M1 | |
| $\theta = 35.3°$ and $144.7°$ (AWRT) | A1 | Both correct. Radians $0.615$, $2.53$ scores A0. |
| | **3** | |7
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } \equiv 1 - \tan ^ { 2 } \theta$.
\item Hence solve the equation $\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } = 2 \tan ^ { 4 } \theta$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q7 [5]}}