## Question 3:

### Part (i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} = \frac{5}{\sin x}$ | B1 | |
| Uses common denominator to give $2\sin^2 x - \cos^2 x = 5\cos x$ | M1 | |
| Replaces $\sin^2 x$ by $(1-\cos^2 x)$ to give $2(1-\cos^2 x)-\cos^2 x = 5\cos x$ | M1 | |
| Obtains $3\cos^2 x + 5\cos x - 2 = 0$ $(a=3,\ b=5,\ c=-2)$ | A1 | |

### Part (i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $3\cos^2 x + 5\cos x - 2 = 0$ to give $\cos x =$ | M1 | |
| $\cos x = \frac{1}{3}$ only (rejects $\cos x = -2$) | A1 | |
| So $x = 1.23$ or $5.05$ | dM1 A1 | |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta + \cot\theta \equiv \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}$ | B1 | Or: $\tan\theta + \cot\theta \equiv \tan\theta + \frac{1}{\tan\theta}$ |
| $\equiv \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}$ | M1 | Or: $\equiv \frac{\tan^2\theta+1}{\tan\theta}$ |
| $\equiv \frac{2}{\sin 2\theta}$ | M1 | Or: $\equiv \frac{1}{\cos^2\theta \times \frac{\sin\theta}{\cos\theta}} \equiv \frac{2}{\sin 2\theta}$ |
| $\equiv 2\cosec 2\theta$ (so $\lambda = 2$) | A1 | |

# Question (i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\tan x = \frac{\sin x}{\cos x}$, $\cot x = \frac{\cos x}{\sin x}$, $\cosec x = \frac{1}{\sin x}$ to write equation in terms of $\cos x$ and $\sin x$ | B1 | Condone $5\cosec x = \frac{1}{5\sin x}$ if intention clear. Alternatively uses $\cot x = \frac{1}{\tan x}$ and $\cosec x = \frac{1}{\sin x}$ to write in terms of $\tan x$ and $\sin x$ |
| Uses common denominator and cross multiples, or multiplies each term by $\sin x \cos x$ to achieve form $A\sin^2 x \pm B\cos^2 x = C\cos x$ | M1 | Alternatively multiplies by $\tan x$ to achieve $A\tan^2 x \pm B = C\sec x$ |
| Uses $\sin^2 x = 1 - \cos^2 x$ to form quadratic in $\cos x$. In alternative uses $\tan^2 x = \sec^2 x - 1$ then $\sec x = \frac{1}{\cos x}$ | M1 | |
| Obtains $\pm K(3\cos^2 x + 5\cos x - 2) = 0$ where $a=3$, $b=5$, $c=-2$ | A1 | |

# Question (i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses standard method to solve quadratic in $\cos x$ from (i)(a) OR $\sec x$ | M1 | |
| $\cos x = \frac{1}{3}$ only | A1 | Do not need to see $-2$ rejected |
| Uses arccos on their value to obtain at least one answer | dM1 | Dependent on previous M. May be implied by one correct answer |
| Both values correct to 3sf: $x = 1.23$ and $5.05$ | A1 | Ignore solutions outside range. Extra solutions score A0. Answers in degrees score A0 |

# Question (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses definition of $\cot$ with matching expression for $\tan$: $\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}$, $\frac{\sin\theta}{\cos\theta} + \frac{1}{\frac{\sin\theta}{\cos\theta}}$, $\tan\theta + \frac{1}{\tan\theta}$ | B1 | Condone miscopy on sign e.g. allow $\tan\theta - \frac{1}{\tan\theta}$ |
| Uses common denominator, writing as single fraction. Denominator must be correct and one term adapted | M1 | In example 2, expression would need to be inverted |
| Uses $\sin^2\theta + \cos^2\theta = 1$ **and** $\sin 2\theta = 2\sin\theta\cos\theta$ to achieve form $\frac{\lambda}{\sin 2\theta}$ | M1 | Alternatively uses $1+\tan^2\theta = \sec^2\theta = \frac{1}{\cos^2\theta}$, $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin 2\theta = 2\sin\theta\cos\theta$. A line of $\frac{1}{\sin\theta\cos\theta}$ followed by $\lambda = \frac{1}{2}$ or $2$ implies this mark |
| Achieves printed answer with no errors | A1 | Allow different variable if used consistently |