| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Moderate -0.3 This is a straightforward application of the Pythagorean identity cosec²x = 1 + cot²x to convert to quadratic form, followed by solving a quadratic and taking reciprocals. The algebraic manipulation is routine and the identity is standard C3 content, making this slightly easier than average but still requiring multiple steps. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(10\csc^2 x = 16 - 11\cot x\); use \(\csc^2 x = 1 + \cot^2 x\) | M1 | |
| \(10(1+\cot^2 x) = 16 - 11\cot x \Rightarrow 10\cot^2 x + 11\cot x - 6 = 0\) | A1 | Shown |
| Answer | Marks |
|---|---|
| \((5\cot x - 2)(2\cot x + 3) = 0\) | M1 |
| \(\cot x = \frac{2}{5}\) or \(\cot x = -\frac{3}{2}\) | A1 |
| \(\tan x = \frac{5}{2}\) or \(\tan x = -\frac{2}{3}\) | M1 A1 |
# Question 5:
## Part (a)
| $10\csc^2 x = 16 - 11\cot x$; use $\csc^2 x = 1 + \cot^2 x$ | M1 | |
| $10(1+\cot^2 x) = 16 - 11\cot x \Rightarrow 10\cot^2 x + 11\cot x - 6 = 0$ | A1 | Shown |
## Part (b)
| $(5\cot x - 2)(2\cot x + 3) = 0$ | M1 | |
| $\cot x = \frac{2}{5}$ or $\cot x = -\frac{3}{2}$ | A1 | |
| $\tan x = \frac{5}{2}$ or $\tan x = -\frac{2}{3}$ | M1 A1 | |
---5
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$10 \operatorname { cosec } ^ { 2 } x = 16 - 11 \cot x$$
can be written in the form
$$10 \cot ^ { 2 } x + 11 \cot x - 6 = 0$$
\item Hence, given that $10 \operatorname { cosec } ^ { 2 } x = 16 - 11 \cot x$, find the possible values of $\tan x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q5 [5]}}