C3 June 2006 Q6
6.
(a) Using \(\sin^2\theta + \cos^2\theta \equiv 1\), show that \(\operatorname{cosec}^2\theta - \cot^2\theta \equiv 1\). (2)
(b) Hence, or otherwise, prove that\[\operatorname{cosec}^4\theta - \cot^4\theta \equiv \operatorname{cosec}^2\theta + \cot^2\theta.\] (2)
(c) Solve, for \(90^\circ \lt \theta \lt 180^\circ\),\[\operatorname{cosec}^4\theta - \cot^4\theta = 2 - \cot\theta.\] (6)
| Scheme | Marks |
|---|---|
| Dividing \(\sin^2\theta + \cos^2\theta \equiv 1\) by \(\sin^2\theta\) to give \(\dfrac{\sin^2\theta}{\sin^2\theta} + \dfrac{\cos^2\theta}{\sin^2\theta} \equiv \dfrac{1}{\sin^2\theta}\) | M1 |
| Completion: \(1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta \Rightarrow \operatorname{cosec}^2\theta - \cot^2\theta \equiv 1\) AG | A1* |
| (2) |
| Scheme | Marks |
|---|---|
| \(\operatorname{cosec}^4\theta - \cot^4\theta \equiv (\operatorname{cosec}^2\theta - \cot^2\theta)(\operatorname{cosec}^2\theta + \cot^2\theta)\) | M1 |
| \(\equiv (\operatorname{cosec}^2\theta + \cot^2\theta)\) using (a) AG | A1* |
| (2) |
Notes
(i) Using LHS \(= (1 + \cot^2\theta)^2 - \cot^4\theta\), using (a) & elim. \(\cot^4\theta\) M1, conclusion {using (a) again} A1*
(ii) Conversion to sines and cosines: needs \(\dfrac{(1 - \cos^2\theta)(1 + \cos^2\theta)}{\sin^4\theta}\) for M1
| Scheme | Marks |
|---|---|
| Using (b) to form \(\operatorname{cosec}^2\theta + \cot^2\theta \equiv 2 - \cot\theta\) | M1 |
| Forming quadratic in \(\cot\theta\) \(\Rightarrow 1 + \cot^2\theta + \cot^2\theta \equiv 2 - \cot\theta\) {using (a)} | M1 |
| \(2\cot^2\theta + \cot\theta - 1 = 0\) | A1 |
| Solving: \((2\cot\theta - 1)(\cot\theta + 1) = 0\) to \(\cot\theta =\) | M1 |
| \(\left(\cot\theta = \tfrac{1}{2}\right)\) or \(\cot\theta = -1\) | A1 |
| \(\theta = 135^\circ\) (or correct value(s) for candidate dep. on 3Ms) | A1ft |
| (6) | |
| (10 marks) |
Notes
Ignore solutions outside range
Extra “solutions” in range loses A1ft, but candidate may possibly have more than one “correct” solution.