12 Show that \(\beta = \arctan \left( \frac { 1 } { 3 } \right)\), as given in line 15 .
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Question 12:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = 1 \Rightarrow \frac{\frac{1}{2} + \tan\beta}{1 - \frac{1}{2}\tan\beta} = 1\) M1
1.1a — Use of \(\tan(\alpha+\beta)\). Accept valid alternative solutions
\(\frac{1}{2} + \tan\beta = 1 - \frac{1}{2}\tan\beta\) M1
1.1 — Rearranging
\(1.5\tan\beta = 0.5 \Rightarrow \tan\beta = \frac{0.5}{1.5} = \frac{1}{3} \Rightarrow \beta = \arctan\left(\frac{1}{3}\right)\) E1
2.1 — Convincing completion. AG. Approximate solutions with decimal angles do not score
[3]
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## Question 12:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = 1 \Rightarrow \frac{\frac{1}{2} + \tan\beta}{1 - \frac{1}{2}\tan\beta} = 1$ | M1 | 1.1a — Use of $\tan(\alpha+\beta)$. Accept valid alternative solutions |
| $\frac{1}{2} + \tan\beta = 1 - \frac{1}{2}\tan\beta$ | M1 | 1.1 — Rearranging |
| $1.5\tan\beta = 0.5 \Rightarrow \tan\beta = \frac{0.5}{1.5} = \frac{1}{3} \Rightarrow \beta = \arctan\left(\frac{1}{3}\right)$ | E1 | 2.1 — Convincing completion. AG. Approximate solutions with decimal angles do not score |
| | **[3]** | |
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12 Show that $\beta = \arctan \left( \frac { 1 } { 3 } \right)$, as given in line 15 .
\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q12 [3]}}