## Question 8:

**Part (a):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan(A+B) = \dfrac{\sin(A+B)}{\cos(A+B)} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}$ | M1 A1 | M1 uses $\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}$ with $sinAcosB \pm cosAsinB$ form; accept incorrect signs |
| $\div \cos A\cos B$: $= \dfrac{\dfrac{\sin A}{\cos A}+\dfrac{\sin B}{\cos B}}{1-\dfrac{\sin A\sin B}{\cos A\cos B}}$ | M1 | Divide both numerator and denominator by $\cos A\cos B$ |
| $= \dfrac{\tan A + \tan B}{1-\tan A\tan B}$ | A1* | Given/cso; both sides must be seen or implied |

**Part (b):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\!\left(\theta+\dfrac{\pi}{6}\right) = \dfrac{\tan\theta + \tan\frac{\pi}{6}}{1-\tan\theta\tan\frac{\pi}{6}}$ | M1 | Use part (a) with $A=\theta$, $B=\frac{\pi}{6}$; accept sign slips |
| $= \dfrac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \tan\theta\cdot\frac{1}{\sqrt{3}}}$ | M1 | Uses $\tan\!\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$ in both numerator and denominator |
| $= \dfrac{\sqrt{3}\tan\theta + 1}{\sqrt{3} - \tan\theta}$ | A1* | Given/cso; all steps correct with no unreasonable jumps |

**Part (c):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\!\left(\theta+\dfrac{\pi}{6}\right) = \tan(\pi-\theta)$ | M1 | Use identity from (b); accept sign slips |
| $\theta + \dfrac{\pi}{6} = \pi - \theta$ | dM1 | Writes equation giving one value of $\theta$; dependent on first M1 |
| $\theta = \dfrac{5}{12}\pi$ | ddM1 A1 | Solves equation in $\theta$; must end with $\theta=$; cso |
| $\tan\!\left(\theta+\dfrac{\pi}{6}\right) = \tan(2\pi-\theta)$ | dddM1 | Equation producing second value |
| $\theta = \dfrac{11}{12}\pi$ | A1 | cso; accept $\frac{5}{12}\pi$ and $\frac{11}{12}\pi$ with no extra solutions in range |

## Question (Part a) - Proof of tan(A+B) identity:

**Main Method:**

| $\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB} = \frac{\sin(A+B)}{\cos(A+B)} = \tan(A+B)$ | A1 | All three expressions must be seen or implied. Given answer - correctly completes proof |

**Alternative (starting from both sides):**

| Uses correct identities for both $\tan A$ and $\tan B$ in rhs expression | M1 | Accept only errors in signs |

| $\frac{\tan A+\tan B}{1-\tan A\tan B} = \frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{1-\frac{\sin A\sin B}{\cos A\cos B}}$ | A1 | |

| Multiplies both numerator and denominator by $\cos A\cos B$ | M1 | Can be stated or implied by working; at least one term modified on both numerator and denominator |

| Completes proof: $\tan(A+B) = \frac{\sin(A+B)}{\cos(A+B)} = \frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}$ | A1 | Must state lhs equals rhs **or** "hence proven" - statement of closure required |

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## Question (Part b) - Alternative from sin and cos:

| Writes $\tan\left(\theta+\frac{\pi}{6}\right) = \frac{\sin(\theta+\frac{\pi}{6})}{\cos(\theta+\frac{\pi}{6})} = \frac{\sin\theta\cos\frac{\pi}{6}+\cos\theta\sin\frac{\pi}{6}}{\cos\theta\cos\frac{\pi}{6}-\sin\theta\sin\frac{\pi}{6}}$ | M1 | |

| Uses $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$ on both numerator and denominator, and divides both by $\cos\theta$ to produce identity in $\tan\theta$ | M1 | |

| As in original scheme | A1 | |

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## Question (Part c) - Solving equation:

**Alternative solution starting with** $1+\sqrt{3}\tan\theta = (\sqrt{3}-\tan\theta)\tan(\pi-\theta)$:

| Let $\tan\theta = t$; expands to $1+\sqrt{3}t=(\sqrt{3}-t)(-t)$, giving $t^2-2\sqrt{3}t-1=0$ | M1 | Expand $\tan(\pi-\theta)$ by correct compound angle identity; substitute $\tan\pi=0$ |

| Collect terms to produce 3-term quadratic in $\tan\theta$ | dM1 | |

| $t = \frac{2\sqrt{3}\pm\sqrt{(12+4)}}{2} = \sqrt{3}\pm 2$ | ddM1 | Correct use of quadratic formula; exact solutions; all previous marks must be scored |

| $\theta = \frac{5\pi}{12}$, $\frac{11\pi}{12}$ | dddM1 | All 3 previous marks must have been scored; produces two exact values for $\theta$ |

| One solution $\frac{5}{12}\pi$ (accept $\frac{\pi}{2.4}$) or $\frac{11}{12}\pi$ | A1 | Must find an exact surd |

| Both solutions $\frac{5}{12}\pi$ and $\frac{11}{12}\pi$, no extra solutions inside range; ignore extra solutions outside range | A1 | |

> **Special case:** Candidates writing $\tan(\pi-\theta)=\tan(\pi)-\tan(\theta)=-\tan(\theta)$ lose first mark but can potentially score others.
> **Solutions in degrees:** Lose first correct mark that would have been scored - usually 75°