| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Compound angle with reciprocal functions |
| Difficulty | Standard +0.8 This question requires using the identity sec²α = 1 + tan²α to derive a quadratic, solving it for exact values, then applying compound angle formulas with reciprocal trig functions. While systematic, it combines multiple non-trivial steps (identity manipulation, exact value arithmetic, compound angle work with cot) beyond standard textbook exercises, placing it moderately above average difficulty. |
| 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^21.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\sec^2\alpha = 1 + \tan^2\alpha\) | B1 | |
| Confirm \(3\tan^2\alpha + 4\tan\alpha - 4 = 0\) | B1 | |
| Solve quadratic equation for \(\tan\alpha\) | M1 | |
| Obtain, finally, \(\tan\alpha = \dfrac{2}{3}\) only | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply \(\tan(\alpha+\beta) = \dfrac{1}{6}\) | B1 | |
| State \(\dfrac{\frac{2}{3} + \tan\beta}{1 - \frac{2}{3}\tan\beta} = \dfrac{1}{6}\), following their value of \(\tan\alpha\) | B1\(\sqrt{}\) | |
| Solve equation of form \(\dfrac{a+bt}{c+dt}\) for \(t\) | M1 | |
| Obtain \(\tan\beta = -\dfrac{9}{20}\) | A1 | |
| Conclude with \(\cot\beta = -\dfrac{20}{9}\) or exact equivalent | A1 | [5] |
## Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\sec^2\alpha = 1 + \tan^2\alpha$ | B1 | |
| Confirm $3\tan^2\alpha + 4\tan\alpha - 4 = 0$ | B1 | |
| Solve quadratic equation for $\tan\alpha$ | M1 | |
| Obtain, finally, $\tan\alpha = \dfrac{2}{3}$ only | A1 | [4] |
## Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $\tan(\alpha+\beta) = \dfrac{1}{6}$ | B1 | |
| State $\dfrac{\frac{2}{3} + \tan\beta}{1 - \frac{2}{3}\tan\beta} = \dfrac{1}{6}$, following their value of $\tan\alpha$ | B1$\sqrt{}$ | |
| Solve equation of form $\dfrac{a+bt}{c+dt}$ for $t$ | M1 | |
| Obtain $\tan\beta = -\dfrac{9}{20}$ | A1 | |
| Conclude with $\cot\beta = -\dfrac{20}{9}$ or exact equivalent | A1 | [5] |7 The angle $\alpha$ lies between $0 ^ { \circ }$ and $90 ^ { \circ }$ and is such that
$$2 \tan ^ { 2 } \alpha + \sec ^ { 2 } \alpha = 5 - 4 \tan \alpha$$
(i) Show that
$$3 \tan ^ { 2 } \alpha + 4 \tan \alpha - 4 = 0$$
and hence find the exact value of $\tan \alpha$.\\
(ii) It is given that the angle $\beta$ is such that $\cot ( \alpha + \beta ) = 6$. Without using a calculator, find the exact value of $\cot \beta$.
\hfill \mbox{\textit{CAIE P2 2014 Q7 [9]}}