| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Given one function find others |
| Difficulty | Moderate -0.8 This question tests standard trigonometric identities and relationships that are routine for A-level students. Part (i) uses the tan(π-x) identity, part (ii) uses the complementary angle relationship (or tan(π/2-x) = cot(x) = 1/tan(x)), and part (iii) requires constructing a right triangle from tan(x)=k to find sin(x)=k/√(1+k²). All three parts involve direct application of well-practiced techniques with no problem-solving or novel insight required, making this easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan(\pi - x) = -k\) | B1 [1] | co. www Mark final answers |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\left(\frac{\pi}{2} - x\right) = \frac{1}{k}\) | B1 [1] | co. www |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin x = \frac{k}{\sqrt{1+k^2}}\) from 90° triangle | M1 A1 [2] | Any valid method – 90° triangle or formulae |
## Question 1:
**Part (i):**
$\tan(\pi - x) = -k$ | B1 [1] | co. www Mark final answers
**Part (ii):**
$\tan\left(\frac{\pi}{2} - x\right) = \frac{1}{k}$ | B1 [1] | co. www
**Part (iii):**
$\sin x = \frac{k}{\sqrt{1+k^2}}$ from 90° triangle | M1 A1 [2] | Any valid method – 90° triangle or formulae
---1 The acute angle $x$ radians is such that $\tan x = k$, where $k$ is a positive constant. Express, in terms of $k$,\\
(i) $\tan ( \pi - x )$,\\
(ii) $\tan \left( \frac { 1 } { 2 } \pi - x \right)$,\\
(iii) $\sin x$.
\hfill \mbox{\textit{CAIE P1 2010 Q1 [4]}}