| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (i) requires algebraic manipulation of trig fractions using standard identities (Pythagorean identity), which is routine for P1 level. Part (ii) applies the proven identity to solve a quadratic in tan θ with restricted domain—straightforward once the identity is established. This is a standard two-part question testing technique rather than insight. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) LHS \(= \frac{\sin^2\theta - (1 - \cos\theta)}{(1 - \cos\theta)\sin\theta}\) cao | B1 | Put over common denominator |
| \(= \frac{1 - \cos^2\theta - 1 - \cos\theta}{(1 - \cos\theta)\sin\theta}\) | M1 | Use \(s^2 = 1 - c^2\) oe |
| \(= \frac{\cos\theta(1 - \cos\theta)}{(1 - \cos\theta)\sin\theta}\) | M1 | Correct factorisation from line 2 |
| \(= \frac{1}{\tan\theta}\) | A1 [4] | AG |
| (ii) \(\tan\theta = (\pm)\frac{1}{2}\) | M1 | |
| \(26.6°, 153.4°\) | A1A1\(\checkmark\) [3] | Ft for \(180 - 1^{\text{st}}\) answer |
(i) LHS $= \frac{\sin^2\theta - (1 - \cos\theta)}{(1 - \cos\theta)\sin\theta}$ cao | B1 | Put over common denominator
$= \frac{1 - \cos^2\theta - 1 - \cos\theta}{(1 - \cos\theta)\sin\theta}$ | M1 | Use $s^2 = 1 - c^2$ oe
$= \frac{\cos\theta(1 - \cos\theta)}{(1 - \cos\theta)\sin\theta}$ | M1 | Correct factorisation from line 2
$= \frac{1}{\tan\theta}$ | A1 [4] | AG
(ii) $\tan\theta = (\pm)\frac{1}{2}$ | M1 |
$26.6°, 153.4°$ | A1A1$\checkmark$ [3] | Ft for $180 - 1^{\text{st}}$ answer9 (i) Prove the identity $\frac { \sin \theta } { 1 - \cos \theta } - \frac { 1 } { \sin \theta } \equiv \frac { 1 } { \tan \theta }$.\\
(ii) Hence solve the equation $\frac { \sin \theta } { 1 - \cos \theta } - \frac { 1 } { \sin \theta } = 4 \tan \theta$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2014 Q9 [7]}}