# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $x=\arctan A$, $y=\arctan B$; $\tan(x-y)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$ | B1 | States tan subtraction formula |
| $\tan(x-y)=\dfrac{A-B}{1+AB} \Rightarrow x-y=\arctan\left(\dfrac{A-B}{1+AB}\right)$ | M1 | Uses $\tan(x-y)=\frac{A-B}{1+AB}$ and takes arctan |
| $\arctan A - \arctan B = \arctan\left(\dfrac{A-B}{1+AB}\right)$ | A1* | Given result; correct conclusion stated |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A=r+2,\ B=r \Rightarrow \dfrac{A-B}{1+AB} = \dfrac{r+2-r}{1+(r+2)r} = \dfrac{2}{\ldots}$ | M1 | Substitutes $A=r+2$, $B=r$ into result from (a) |
| $= \dfrac{2}{r^2+2r+1} = \dfrac{2}{(1+r)^2}$ | A1* | Given answer, correct simplification |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\sum_{r=1}^{n}\arctan\!\left(\dfrac{2}{(1+r)^2}\right) = \sum_{r=1}^{n}(\arctan(r+2)-\arctan(r))$ | M1 | Uses result from (b) to write as telescoping sum |
| $= (\arctan 3-\arctan 1)+(\arctan 4-\arctan 2)+(\arctan 5-\arctan 3)+\ldots$ $+(\arctan(n+1)-\arctan(n-1))+(\arctan(n+2)-\arctan n)$ | A1 | Writes out terms showing telescoping pattern |
| $= \arctan(n+2)+\arctan(n+1)-\arctan 2-\arctan 1$ | M1 | Identifies surviving terms after cancellation |
| $= \arctan(n+2)+\arctan(n+1)-\arctan 2-\dfrac{\pi}{4}$ | A1 | $\arctan 1 = \frac{\pi}{4}$ substituted correctly |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| As $n\to\infty$, $\arctan n \to \dfrac{\pi}{2}$ | M1 | States correct limit of arctan |
| $\displaystyle\sum_{r=1}^{\infty}\arctan\!\left(\dfrac{2}{(1+r)^2}\right) = \dfrac{\pi}{2}+\dfrac{\pi}{2}-\arctan 2-\dfrac{\pi}{4} = \dfrac{3\pi}{4}-\arctan 2$ | A1 | Correct final answer |

# Question 6 (Compound Angle / arctan series):

## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Correct statement or use of compound angle formula with consistent variables of $x$ and $y$ or $\arctan A$ and $\arctan B$ | B1 | Can be either way round |
| Attempts to apply tan or arctan on appropriate identity with either $x$ and $y$ or $\arctan A$ and $\arctan B$; should have $\frac{\tan x \pm \tan y}{1 \pm \tan x \tan y}$ | M1 | oe with arctans or $A$'s and $B$'s |
| Must have scored B and M marks; replaces $\tan x$ and $\tan y$ by $A$ and $B$ with fully correct work and conclusion, no erroneous statements | A1* | Conclusion must include reference to identity being true |

**Working in reverse note:**
Let $x = \arctan A$, $y = \arctan B$:
$$\arctan A - \arctan B = \arctan\left(\frac{A-B}{1+AB}\right) \Leftrightarrow \tan(x-y) = \frac{A-B}{1+AB}$$ Scores M1
$$\Leftrightarrow \tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$$ Scores B1

## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Substitutes $A = r+2$ and $B = r$ and simplifies numerator to 2 | M1 | May be implied |
| Expands denominator (must be seen) and factorises to given result, no errors | A1* | Must be seen |

## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Applies result of (a) to the series | M1 | Allow if different $A$ and $B$ due to error |
| At least first three and final two brackets of terms correctly written out | A1 | Must be clear enough to show cancelling |
| Extracts the non-cancelling terms | M1 | |
| Correct result with no errors — must see $\arctan 1$ before reaching $\frac{\pi}{4}$ | A1 | Insufficient terms for first A is not an error; M1A0M1A1 possible |

## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| Identifies value $\arctan$ tends towards as $n$ increases | M1 | Need not see limits, as long as value identified |
| Correct answer | A1 | |

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