# Question 7(a) - Trig Proof

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{cosec}\,2x + \cot 2x = \frac{1}{\sin 2x} + \frac{\cos 2x}{\sin 2x}$ | M1 | Writing $\text{cosec}\,2x = \frac{1}{\sin 2x}$ and $\cot 2x = \frac{\cos 2x}{\sin 2x}$ or $\frac{1}{\tan 2x}$ |
| $= \frac{1 + \cos 2x}{\sin 2x}$ | M1 | Writing as single fraction $\frac{a+b}{c}$, denominator correct |
| $= \frac{1 + 2\cos^2 x - 1}{2\sin x \cos x} = \frac{2\cos^2 x}{2\sin x \cos x}$ | M1 A1 | Uses double angle formulae to get form with no addition/subtraction signs. Accept $\frac{2\cos^2 x}{2\sin x \cos x}$ |
| $= \frac{\cos x}{\sin x} = \cot x$ | A1* | Completes proof by cancelling, using $\frac{\cos x}{\sin x} = \cot x$ or $\frac{1}{\tan x} = \cot x$ |

# Question 7(b) - Trig Equation

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{cosec}(4\theta+10°) + \cot(4\theta+10°) = \sqrt{3} \Rightarrow \cot(2\theta \pm ...°) = \sqrt{3}$ | M1 | Use result from (a) with $2x = 4\theta+10$. Accept solution from $\cot(2x\pm...°)=\sqrt{3}$ |
| $2\theta \pm ... = 30° \Rightarrow \theta = 12.5°$ | dM1, A1 | Proceeds using $\tan.. = \frac{1}{\cot..}$, $\arctan\left(\frac{1}{\sqrt{3}}\right) = 30°$ |
| $2\theta \pm ... = 180 + PV° \Rightarrow \theta = ..°$ | dM1 | Correct method for second solution, dependent on first M1 only |
| $\theta = 102.5°$ | A1 | Ignore solutions outside range. Withhold for extra solutions within range |

## Question 7(a) Alt 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cosec 2x + \cot 2x = \frac{1}{\sin 2x} + \frac{1}{\tan 2x}$ | 1st M1 | |
| $= \frac{1}{2\sin x \cos x} + \frac{1-\tan^2 x}{2\tan x}$ | 2nd M1 | Combining fractions with common denominator |
| $= \frac{\tan x + (1-\tan^2 x)\sin x \cos x}{2\sin x \cos x \tan x}$ or $= \frac{2\tan x + 2(1-\tan^2 x)\sin x \cos x}{4\sin x \cos x \tan x}$ | 2nd M1 | |
| $= \frac{\tan x + \sin x \cos x - \tan^2 x \sin x \cos x}{2\sin x \cos x \tan x}$ | | |
| $= \frac{\tan x + \sin x \cos x - \tan x \sin^2 x}{2\sin x \cos x \tan x}$ | | |
| $= \frac{\tan x(1-\sin^2 x) + \sin x \cos x}{2\sin x \cos x \tan x}$ | | |
| $= \frac{\tan x \cos^2 x + \sin x \cos x}{2\sin x \cos x \tan x}$ | | |
| $= \frac{\sin x \cos x + \sin x \cos x}{2\sin x \cos x \tan x}$ | | |
| $= \frac{2\sin x \cos x}{2\sin x \cos x \tan x}$ oe | 3rd M1A1 | |
| $= \frac{1}{\tan x} = \cot x$ | A1* | (5) |

## Question 7(a) Alt 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cosec 2x + \cot 2x = \cot x \Leftrightarrow \frac{1}{\sin 2x} + \frac{1}{\tan 2x} = \frac{1}{\tan x}$ | 1st M1 | |
| $\Leftrightarrow \tan 2x \tan x + \sin 2x \tan x = \sin 2x \tan 2x$ | 2nd M1 | |
| $\Leftrightarrow \frac{2\tan x}{1-\tan^2 x} \times \tan x + 2\sin x \cos x \times \frac{\sin x}{\cos x} = 2\sin x \cos x \times \frac{2\tan x}{1-\tan^2 x}$ | | |
| $\Leftrightarrow \frac{2\tan^2 x}{1-\tan^2 x} + 2\sin^2 x = \frac{4\sin^2 x}{1-\tan^2 x}$ | | |
| $\times(1-\tan^2 x) \Leftrightarrow 2\tan^2 x + 2\sin^2 x(1-\tan^2 x) = 4\sin^2 x$ | 3rd M1 | |
| $\Leftrightarrow 2\tan^2 x - 2\sin^2 x \tan^2 x = 2\sin^2 x$ | A1 | |
| $\Leftrightarrow 2\tan^2 x(1-\sin^2 x) = 2\sin^2 x$ | | |
| $\div 2\tan^2 x \Leftrightarrow 1 - \sin^2 x = \cos^2 x$ | | |
| As this is true, initial statement is true | A1* | (5) |

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