| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove algebraic trigonometric identity |
| Difficulty | Standard +0.3 Part (a) is a standard reciprocal trig identity proof requiring conversion to sin/cos and use of the double angle formula - routine for Further Maths students. Part (b) is a simple application noting that cosec 2θ ≥ 1, making this slightly easier than average even for A-level. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\theta + \cot\theta \equiv \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\) | M1 | Writes \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and \(\cot\theta = \frac{\cos\theta}{\sin\theta}\) |
| \(\equiv \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}\) | A1 | Correct intermediate answer |
| \(\equiv \frac{1}{\frac{1}{2}\sin 2\theta}\) | M1 | Uses double angle formula \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(\equiv 2\cosec 2\theta\) | A1* | Completes proof with no errors (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(\tan\theta + \cot\theta = 1 \Rightarrow \sin 2\theta = 2\) AND no real solutions as \(-1 \leqslant \sin 2\theta \leqslant 1\) | B1 | Scored for sight of \(\sin 2\theta = 2\) and a reason why no real solutions exist; possible reasons: \(-1 \leqslant \sin 2\theta \leqslant 1\) therefore \(\sin 2\theta \neq 2\), or \(2\theta = \arcsin 2\) which has no answers as \(-1 \leqslant \sin 2\theta \leqslant 1\) |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta + \cot\theta \equiv \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}$ | M1 | Writes $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\cot\theta = \frac{\cos\theta}{\sin\theta}$ |
| $\equiv \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}$ | A1 | Correct intermediate answer |
| $\equiv \frac{1}{\frac{1}{2}\sin 2\theta}$ | M1 | Uses double angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $\equiv 2\cosec 2\theta$ | A1* | Completes proof with no errors (given answer) |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $\tan\theta + \cot\theta = 1 \Rightarrow \sin 2\theta = 2$ AND no real solutions as $-1 \leqslant \sin 2\theta \leqslant 1$ | B1 | Scored for sight of $\sin 2\theta = 2$ and a reason why no real solutions exist; possible reasons: $-1 \leqslant \sin 2\theta \leqslant 1$ therefore $\sin 2\theta \neq 2$, or $2\theta = \arcsin 2$ which has no answers as $-1 \leqslant \sin 2\theta \leqslant 1$ |
---\begin{enumerate}
\item (a) Prove that
\end{enumerate}
$$\tan \theta + \cot \theta \equiv 2 \operatorname { cosec } 2 \theta , \quad \theta \neq \frac { n \pi } { 2 } , n \in \mathbb { Z }$$
(b) Hence explain why the equation
$$\tan \theta + \cot \theta = 1$$
does not have any real solutions.\\
\hfill \mbox{\textit{Edexcel Paper 1 Q9 [5]}}