| 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) is a standard algebraic manipulation of trig identities requiring common denominator work and Pythagorean identity application. Part (ii) follows directly by substituting the proven identity to get tan θ = -2, then solving a routine inverse tangent problem with quadrant consideration. This is typical P1 material with straightforward 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 |
|---|---|---|
| (i) \(\frac{1}{\cos\theta} + \frac{\cos\theta}{1+\sin\theta} = \tan\theta\). | M1 | Correct addition of fractions |
| LHS = \(\frac{1+s-c^2}{c(1+s)} = \frac{s^2+s}{c(1+s)} = \frac{s}{c}\) | M1M1 | Use of \(s^2+c^2=1\). \((1+s)\)cancelled. |
| = \(\tan\theta\) | A1 [4] | → answer given. |
| (ii) → \(\tan\theta + 2 = 0\) ie \(\tan\theta = -2\) → \(\theta = 116.6°\) or \(296.6°\) | M1 A1 A1✓ [3] | Uses part (i). Allow \(\tan\theta=±2\) Co. ✓ for 180°+ and no other solutions in the range. |
**(i)** $\frac{1}{\cos\theta} + \frac{\cos\theta}{1+\sin\theta} = \tan\theta$. | M1 | Correct addition of fractions
LHS = $\frac{1+s-c^2}{c(1+s)} = \frac{s^2+s}{c(1+s)} = \frac{s}{c}$ | M1M1 | Use of $s^2+c^2=1$. $(1+s)$cancelled.
= $\tan\theta$ | A1 [4] | → answer given.
**(ii)** → $\tan\theta + 2 = 0$ ie $\tan\theta = -2$ → $\theta = 116.6°$ or $296.6°$ | M1 A1 A1✓ [3] | Uses part (i). Allow $\tan\theta=±2$ Co. ✓ for 180°+ and no other solutions in the range.5 (i) Prove the identity $\frac { 1 } { \cos \theta } - \frac { \cos \theta } { 1 + \sin \theta } \equiv \tan \theta$.\\
(ii) Solve the equation $\frac { 1 } { \cos \theta } - \frac { \cos \theta } { 1 + \sin \theta } + 2 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2014 Q5 [7]}}