| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| 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 This is a straightforward two-part question requiring standard algebraic manipulation of trig identities followed by a routine equation solve. Part (i) involves expressing tan x as sin x/cos x and simplifying—a textbook exercise. Part (ii) is a direct application leading to cos x = -1/2, giving standard angles. Slightly above average due to the algebraic manipulation required, but no novel insight needed. |
| Spec | 1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\sin x \tan x}{1-\cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)}\) | \(M1\) | Use of \(\tan x = \sin x \div \cos x\) |
| \(= \frac{1-\cos^2 x}{\cos x(1-\cos x)}\) | \(M1\) | Use of \(\sin^2 x = 1 - \cos^2 x\) |
| \(= \frac{(1-\cos x)(1+\cos x)}{\cos x(1-\cos x)} = \frac{1}{\cos x}+1\) | \(M1\) | Realising the need to use difference of 2 squares. Answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{\cos x}+1+2=0 \rightarrow \cos x = -\frac{1}{3}\) | \(M1\) | Uses part (i) with \(\cos x\) as subject |
| \(\rightarrow x = 109.5°\) or \(250.5°\) | \(A1\ A1\sqrt{}\) | co. \(\sqrt{}\) for \(360°\) – \(1^{st}\) answer |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\sin x \tan x}{1-\cos x} = \frac{\sin^2 x}{\cos x(1-\cos x)}$ | $M1$ | Use of $\tan x = \sin x \div \cos x$ |
| $= \frac{1-\cos^2 x}{\cos x(1-\cos x)}$ | $M1$ | Use of $\sin^2 x = 1 - \cos^2 x$ |
| $= \frac{(1-\cos x)(1+\cos x)}{\cos x(1-\cos x)} = \frac{1}{\cos x}+1$ | $M1$ | Realising the need to use difference of 2 squares. Answer given. |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\cos x}+1+2=0 \rightarrow \cos x = -\frac{1}{3}$ | $M1$ | Uses part (i) with $\cos x$ as subject |
| $\rightarrow x = 109.5°$ or $250.5°$ | $A1\ A1\sqrt{}$ | co. $\sqrt{}$ for $360°$ – $1^{st}$ answer |
---4 (i) Prove the identity $\frac { \sin x \tan x } { 1 - \cos x } \equiv 1 + \frac { 1 } { \cos x }$.\\
(ii) Hence solve the equation $\frac { \sin x \tan x } { 1 - \cos x } + 2 = 0$, for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2010 Q4 [6]}}