# Question 3:

## Way 1:

| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{d}{dx}\left(\frac{\sin x}{\cos x-1}\right) = \frac{\cos x(\cos x-1)+\sin^2 x}{(\cos x-1)^2}$ | M1A1 | M1: Correct use of quotient (or product) rule; A1: Correct expression |
| $\frac{dy}{dx} = \frac{1}{1+\left(\frac{\sin x}{\cos x-1}\right)^2}\left(\frac{\cos x(\cos x-1)+\sin^2 x}{(\cos x-1)^2}\right)$ | dM1A1 | dM1: $\frac{1}{1+\left(\frac{\sin x}{\cos x-1}\right)^2}\times$ quotient (or product) rule; must be function of $x$; A1: Correct expression |
| $\frac{dy}{dx} = \frac{(\cos x-1)^2}{(\cos x-1)^2+\sin^2 x}\left(\frac{1-\cos x}{(\cos x-1)^2}\right)=\frac{1}{2}$ | ddM1A1 | ddM1: Attempts to simplify to obtain a constant; must reach a constant; A1: cao |

**Special Case:** Quotient rule used with numerator terms wrong way round and work otherwise correct: award M1A0 and M1A0ddM1A0 if rest of method correct.

## Way 2:

| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{d}{dx}\left(\frac{\sin x}{\cos x-1}\right) = \frac{\cos x(\cos x-1)+\sin^2 x}{(\cos x-1)^2}$ | M1A1 | M1: Correct use of quotient (or product) rule; A1: Correct expression |
| $\tan y = \left(\frac{\sin x}{\cos x-1}\right)\Rightarrow \sec^2 y\frac{dy}{dx}=\frac{\cos x(\cos x-1)+\sin^2 x}{(\cos x-1)^2}$ | | |
| $\frac{dy}{dx}=\frac{1}{1+\left(\frac{\sin x}{\cos x-1}\right)^2}\left(\frac{\cos x(\cos x-1)+\sin^2 x}{(\cos x-1)^2}\right)$ | dM1A1 | dM1: $\frac{1}{1+\left(\frac{\sin x}{\cos x-1}\right)^2}\times$ quotient/product rule; must be function of $x$; A1: Correct expression |
| $\frac{dy}{dx}=\frac{(\cos x-1)^2}{(\cos x-1)^2+\sin^2 x}\left(\frac{1-\cos x}{(\cos x-1)^2}\right)=\frac{1}{2}$ | ddM1A1 | ddM1: Attempts to simplify to obtain a constant; A1: cao |

## Way 3:

| Working/Answer | Marks | Notes |
|---|---|---|
| $\tan y=\left(\frac{\sin x}{\cos x-1}\right)\Rightarrow (\cos x-1)\tan y=\sin x \Rightarrow -\sin x\tan y+(\cos x-1)\sec^2 y\frac{dy}{dx}=\cos x$ | M1A1 | M1: Differentiates implicitly; A1: Correct differentiation |
| $\Rightarrow \frac{-\sin^2 x}{\cos x-1}+(\cos x-1)\left(1+\frac{\sin^2 x}{(\cos x-1)^2}\right)\frac{dy}{dx}=\cos x$ | dM1A1 | dM1: Substitutes for $y$ throughout; A1: Correct equation in terms of $x$ only (and $dy/dx$) |
| $\frac{dy}{dx}=\frac{1}{2}$ | ddM1A1 | ddM1: Attempts to simplify to obtain a constant; A1: cao |

## Way 4:

| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{\sin x}{\cos x-1}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{1-2\sin^2\frac{x}{2}-1}$ | M1A1 | M1: Using the correct double angle formula; A1: Correct expression |
| $=-\cot\frac{x}{2}=-\tan\left(\frac{\pi}{2}\pm\frac{x}{2}\right)=\tan\left(\frac{x}{2}\pm\frac{\pi}{2}\right)$ | dM1A1 | M1: Obtains $\tan$ in terms of $x$; A1: $\tan\left(\frac{x}{2}+\frac{\pi}{2}\right)$ |
| So $y=\arctan\left(\tan\left(\frac{x}{2}\pm\frac{\pi}{2}\right)\right)\Rightarrow\frac{dy}{dx}=\frac{1}{2}$ | ddM1A1 | ddM1: Attempts to simplify to obtain a constant; A1: cao |

**Total: 6 marks**

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