**(i)**

| $\frac{\tan 2x + \tan 32°}{1 - \tan 2x \tan 32°} = 5 \Rightarrow \tan(2x+32°) = 5$ | B1 |
| $\Rightarrow x = \frac{\arctan 5 - 32°}{2}$ | M1 |
| $\Rightarrow x = awrt 23.35°, -66.65°$ | A1A1 |

(4 marks)

**(ii)(a)**

| $\tan(3\theta - 45°) = \frac{\tan 3\theta - \tan 45°}{1 + \tan 45° \tan 3\theta} = \frac{\tan 3\theta - 1}{1 + \tan 3\theta}$ | M1A1* |

(2 marks)

**(ii)(b)**

| $(1+\tan 3\theta)\tan(\theta+28°) = \tan 3\theta - 1$ | B1 |
| $\Rightarrow \tan(\theta + 28°) = \tan(3\theta - 45°)$ | M1A1 |
| $\theta + 28° = 3\theta - 45° \Rightarrow \theta = 36.5°$ | M1A1 |
| $\theta + 28° + 180° = 3\theta - 45° \Rightarrow \theta = 126.5°$ | dM1A1 |

(5 marks)
(11 marks total)

**Guidance notes:**

**(i)**

- B1: Stating or implying by subsequent work $\tan(2x+32°) = 5$
- M1: Scored for the correct order of operations from $\tan(2x \pm 32°) = 5$ to $x = ...$, $x = \frac{\arctan 5 \pm 32°}{2}$. This may be implied by one correct answer
- A1: One of awrt $x = 23.3 / 23.4°, -66.6 / -66.7°$. One dp accuracy required for this penultimate mark.
- A1: Both of $x = awrt 23.35°, -66.65°$ and no other solutions in the range $-90° < x < 90°$

**Using Alt I:**

| B1: tan $2x = awrt 1.06$ | |
| M1: For attempting to make tan $2x$ the subject followed by correct inverse operations to find a value for $x$ | |
| A1A1: (4) | |

**Using Alt 2:**

| B1: $\frac{2\tan x}{1-\tan^2 x} = 5 - 5x$ oe | |
| M1: For attempting to make tan $2x$ the subject followed by correct inverse operations to find a value for $x$ | |
| A1 A1: (4) | |

**(ii)(a)**

- M1: States or implies (just rhs) $\tan(3\theta - 45°) = \frac{\tan 3\theta \pm \tan 45°}{1 \pm \tan 45° \tan 3\theta}$
- A1*: Complete proof with the lhs, the correct identity $\frac{\tan 3\theta - \tan 45°}{1 + \tan 45° \tan 3\theta}$ and either stating that $\tan 45° = 1$ or substituting $\tan 45° = 1$ (which may only be seen on the numerator) and proceeding to given answer. It is possible to work backwards here: $\frac{\tan 3\theta - 1}{1 + \tan 3\theta} = \tan(3\theta - 45°)$ with M1 A1 scored at the end. Do not allow the final A1* if there are errors.

**(ii)(b)**

- B1: Uses (ii)(a) to state or imply that $\tan(\theta + 28°) = \tan(3\theta - 45°)$. Allow this mark for $(1+\tan 3\theta)\tan(\theta+28°)=(1+\tan 3\theta)\tan(3\theta-45°)$
- M1: $\theta + 28° = 3\theta - 45° \Rightarrow \theta = ...$. We have seen two incorrect methods that should be given M0:
  - $\tan(\theta+28°) = \tan(3\theta - 45°) \Rightarrow \tan(3\theta - 45°) - \tan(\theta+28°) = 0 \Rightarrow (3\theta - 45°) - (\theta + 28°) = 0 \Rightarrow \theta = ...$
  - and $\tan 3\theta - \tan 45° = \tan \theta + \tan 28° \Rightarrow 3\theta - 45° = \theta + 28° \Rightarrow \theta = ...$
- A1: $\theta = 36.5°$ oe such as $\frac{73}{2}$
- dM1: A correct method of finding a 2nd solution $\theta + 28° + 180° = 3\theta - 45° \Rightarrow \theta = ...$. The previous M must have been awarded. The method may be implied by their $\theta_1 + 90°$ but only if the previous M was scored.
- A1: $\theta = 36.5°, 126.5°$ oe and no other solutions in the range.

The questions states 'hence' so the minimum expected working is $\tan(\theta+28°) = \tan(3\theta-45°)$. Full marks can be awarded when this point is reached.

**Alternative solution using compound angles:**

From the B1 mark, $\tan(\theta+28°)=\tan(3\theta-45°)$ they proceed to
$$\frac{\sin(\theta+28°)}{\cos(\theta+28°)} = \frac{\sin(3\theta-45°)}{\cos(3\theta-45°)} \Rightarrow \sin((3\theta-45°)-(\theta+28°)) = 0$$ via the compound angle identity

So, M1 is gained for an attempt at one value for $\sin(2\theta-73°)=0$, condoning slips and A1 for $\theta = 36.5°$.

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