| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a two-part question requiring algebraic manipulation of trigonometric expressions and solving. Part (i) involves combining fractions with common denominator (1-cos²θ = sin²θ) and simplifying using tan θ = sin θ/cos θ - standard technique practice. Part (ii) is straightforward once the identity is established: set the simplified form to zero and solve tan θ = cos θ, yielding θ = 45°. While multi-step, it follows predictable patterns without requiring novel insight, making it slightly easier than average. |
| 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 intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks |
|---|---|
| \(\dfrac{(\tan\theta+1)(1-\cos\theta)+(\tan\theta-1)(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}\) soi | M1 |
| \(\dfrac{\tan\theta - \tan\theta\cos\theta + 1 - \cos\theta + \tan\theta - 1 + \tan\theta\cos\theta - \cos\theta}{1-\cos^2\theta}\) www | A1 |
| \(\dfrac{2(\tan\theta - \cos\theta)}{\sin^2\theta}\) www AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((2)(\tan\theta - \cos\theta)(=0) \rightarrow (2)\left(\dfrac{\sin\theta}{\cos\theta} - \cos\theta\right)(=0)\) soi | M1 | Equate numerator to zero and replace \(\tan\theta\) by \(\sin\theta/\cos\theta\) |
| \((2)\left(\sin\theta - (1-\sin^2\theta)\right)(=0)\) | DM1 | Multiply by \(\cos\theta\) and replace \(\cos^2\theta\) by \(1-\sin^2\theta\) |
| \(\sin\theta = 0.618(0)\) soi | A1 | Allow \((\sqrt{5}-1)/2\) |
| \(\theta = 38.2°\) | A1 | Apply penalty \(-1\) for extra solutions in range |
## Question 7(i):
| $\dfrac{(\tan\theta+1)(1-\cos\theta)+(\tan\theta-1)(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}$ soi | M1 | |
|---|---|---|
| $\dfrac{\tan\theta - \tan\theta\cos\theta + 1 - \cos\theta + \tan\theta - 1 + \tan\theta\cos\theta - \cos\theta}{1-\cos^2\theta}$ www | A1 | |
| $\dfrac{2(\tan\theta - \cos\theta)}{\sin^2\theta}$ www AG | A1 | |
## Question 7(ii):
| $(2)(\tan\theta - \cos\theta)(=0) \rightarrow (2)\left(\dfrac{\sin\theta}{\cos\theta} - \cos\theta\right)(=0)$ soi | M1 | Equate numerator to zero and replace $\tan\theta$ by $\sin\theta/\cos\theta$ |
|---|---|---|
| $(2)\left(\sin\theta - (1-\sin^2\theta)\right)(=0)$ | DM1 | Multiply by $\cos\theta$ and replace $\cos^2\theta$ by $1-\sin^2\theta$ |
| $\sin\theta = 0.618(0)$ soi | A1 | Allow $(\sqrt{5}-1)/2$ |
| $\theta = 38.2°$ | A1 | Apply penalty $-1$ for extra solutions in range |
---7 (i) Show that $\frac { \tan \theta + 1 } { 1 + \cos \theta } + \frac { \tan \theta - 1 } { 1 - \cos \theta } \equiv \frac { 2 ( \tan \theta - \cos \theta ) } { \sin ^ { 2 } \theta }$.\\
(ii) Hence, showing all necessary working, solve the equation
$$\frac { \tan \theta + 1 } { 1 + \cos \theta } + \frac { \tan \theta - 1 } { 1 - \cos \theta } = 0$$
for $0 ^ { \circ } < \theta < 90 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q7 [7]}}