## Question 1:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin(r+1)}{\cos(r+1)} - \frac{\sin r}{\cos r} = \frac{\sin(r+1)\cos r - \cos(r+1)\sin r}{\cos(r+1)\cos r} = \frac{\sin(r+1-r)}{\cos(r+1)\cos r}$ | M1 | Applies all relevant correct addition formulae from MF19 for the route chosen and combines into a single fraction. |
| OR $\frac{\sin(r+1)}{\cos(r+1)} - \frac{\sin r}{\cos r} = \frac{\sin r\cos 1 + \cos r\sin 1}{\cos(r+1)} - \frac{\sin r}{\cos r}$ | | |
| $= \frac{\sin r\cos r\cos 1 + \cos^2 r\sin 1 - \sin r\cos r\cos 1 + \sin^2 r\sin 1}{\cos(r+1)\cos r}$ | | |
| $= \frac{\sin 1}{\cos(r+1)\cos r}$ | A1 | SC B2 for this route completely correct to AG. |
| Alternative via $\tan$: $\frac{\tan r + \tan 1}{1 - \tan r\tan 1} - \tan r = \frac{\tan r + \tan 1 - \tan r + \tan^2 r\tan 1}{1-\tan r\tan 1}$ | | |
| $= \frac{\tan 1\sec^2 r}{1-\tan r\tan 1} = \frac{\tan 1}{\cos r(\cos r - \sin r\tan 1)} = \frac{\sin 1}{\cos r(\cos r\cos 1 - \sin r\sin 1)} = \frac{\sin 1}{\cos r\cos(r+1)}$ | | |
| **Total** | **2** | |

### Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\sin 1]\displaystyle\sum_{r=1}^{n} u_r = \tan 2 - \tan 1 + \tan 3 - \tan 2 + \ldots + \tan(n+1) - \tan n$ | M1 A1 | Shows enough terms for cancellation to be clear. There must be three complete terms including first and last. Cancelling may be implied. |
| $\displaystyle\sum_{r=1}^{n} u_r = \frac{\tan(n+1) - \tan 1}{\sin 1}$ | A1 | OE and ISW. |
| **Total** | **3** | |

# Question 1(c):
$\tan(n+1)$ oscillates as $n \to \infty$ so $u_1 + u_2 + u_3 + \ldots$ does not converge. | **B1** | States 'oscillates' or refers to diverging values of $\tan(n+1)$, or states that $\tan(n+1)$ does not tend to a limit.

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