## Question 5(a) Way One (LHS to RHS):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cot^2 x - \tan^2 x \equiv \frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\cos^2 x} \equiv \frac{\cos^4 x - \sin^4 x}{\sin^2 x \cos^2 x}$ | M1 | Changes LHS to $\sin x$ and $\cos x$, attempts single fraction with correct common denominator. Condone errors/slips on numerator |
| $\equiv \frac{(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)}{\sin^2 x \cos^2 x} \equiv \frac{\cos 2x}{\ldots}$ or $\frac{\ldots}{\left(\frac{1}{2}\sin 2x\right)^2}$ | dM1 | Applies either $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x$ on numerator, or $\sin 2x = 2\sin x \cos x$ to denominator (condoning bracketing slip) |
| $\equiv \frac{\cos 2x}{\left(\frac{1}{2}\sin 2x\right)^2}$ | A1 | Applies both of the above correctly to achieve correct expression in terms of $\cos 2x$ and $\sin 2x$ |
| $\equiv 4\frac{\cos 2x}{\sin 2x \sin 2x} \equiv 4\cot 2x \operatorname{cosec} 2x$ * | A1* | Reaches RHS with sufficient working shown. Penalise consistent (not once or twice) use of poor notation on this mark only |

**(4 marks)**

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## Question 5(a) Way Two (RHS to LHS):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\cot 2x \operatorname{cosec} 2x \equiv 4\frac{\cos 2x}{\sin 2x} \times \frac{1}{\sin 2x} \equiv \frac{4(\cos^2 x - \sin^2 x)}{\ldots}$ or $\frac{\ldots}{4\sin^2 x \cos^2 x}$ | M1 | Changes to $\sin 2x$ and $\cos 2x$ or $\tan 2x$ and $\sin 2x$, attempts single angles in $\sin x$ and $\cos x$ |
| $\frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} \equiv \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x} \equiv \operatorname{cosec}^2 x - \sec^2 x$ | dM1A1 | Changes to single angles throughout, splits into 2 separate fractions; correct expression in terms of $\operatorname{cosec} x$ and $\sec x$ |
| $\equiv 1 + \cot^2 x - 1 - \tan^2 x \equiv \cot^2 x - \tan^2 x$ * | A1* | Reaches LHS with sufficient working shown |

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## Question 5(a) Way Three (Working on Both Sides):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\cos^2 x} \equiv 4 \times \frac{\cos 2x}{\sin 2x} \times \frac{1}{\sin 2x}$; $\cos^4 x - \sin^4 x \equiv 4\sin^2 x \cos^2 x \frac{\cos 2x}{\sin 2x} \times \frac{1}{\sin 2x}$ | M1 | Changes to $\sin x$, $\cos x$, $\sin 2x$ and $\cos 2x$, attempts to cross multiply |
| $\cos^4 x - \sin^4 x \equiv 4 \times \sin^2 x \cos^2 x \times \frac{(\cos^2 x - \sin^2 x)}{(2\sin x \cos x)^2}$ | dM1A1 | Applies either $\cos^2 x - \sin^2 x = \cos 2x$ to numerator or $\sin 2x = 2\sin x \cos x$ to denominator; correct identity in terms of $\cos x$ and $\sin x$ |
| $(\cos^4 x - \sin^4 x) \equiv (\cos^2 x - \sin^2 x)$; $(\cos^2 x - \sin^2 x)\overbrace{(\cos^2 x + \sin^2 x)}^{1} \equiv (\cos^2 x - \sin^2 x)$ Hence true | A1* | Reaches point where both sides are equal and makes a minimal comment |

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## Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\cot 2\theta \operatorname{cosec} 2\theta = 2\tan^2\theta \Rightarrow \cot^2\theta - \tan^2\theta = 2\tan^2\theta \Rightarrow \cot^2\theta - 3\tan^2\theta = 0$ | M1 | Uses part (a) and attempts to collect terms. See Appendix III for ways not using part (a). Equations in $\sin\theta$ or $\cos\theta$ where mark awarded for equation in just $\sin\theta$ or $\cos\theta$ |
| $\cot^2\theta - 3\tan^2\theta = 0 \Rightarrow \frac{1}{\tan^2\theta} - 3\tan^2\theta = 0 \Rightarrow \tan^4\theta = \frac{1}{3}$ | A1 | Reaches correct equation in single term, usually $\tan\theta$. Look for $\tan^4\theta = \frac{1}{3}$. Other correct intermediate forms: $2\sin^4\theta + 2\sin^2\theta - 1 = 0$ and $2\cos^4\theta - 6\cos^2\theta + 3 = 0$ |
| $\tan^4\theta = "\frac{1}{3}" \Rightarrow \tan\theta = \pm\sqrt[4]{"\frac{1}{3}"} = "\pm 0.7598\ldots" \Rightarrow \theta = \ldots$ | M1 | Takes the 4th root of their $\frac{1}{3}$ (o.e.) and uses $\tan^{-1}$ to obtain at least one value for $\theta$. For other intermediate forms: $\sin^2\theta = \frac{\sqrt{3}-1}{2} \Rightarrow \sin\theta = \sqrt{\frac{\sqrt{3}-1}{2}} \Rightarrow \theta = \ldots$ |
| $\theta =$ awrt $0.65$, $-0.65$ | A1A1 | Either awrt $0.65$ or awrt $-0.65$. Allow either answer in degrees, so awrt $\pm 37.2°$. Both answers in radians, awrt $\pm 0.65$, and no extras in range |

**(5 marks) Total 9**

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