C3 June 2008 Q5
5.
(a) Given that \(\sin^2\theta + \cos^2\theta \equiv 1\), show that \(1 + \cot^2\theta \equiv \mathrm{cosec}^2\,\theta\). (2)
(b) Solve, for \(0 \leqslant \theta < 180^\circ\), the equation\[2\cot^2\theta - 9\,\mathrm{cosec}\,\theta = 3,\]giving your answers to 1 decimal place. (6)
| Scheme | Marks |
|---|---|
| \(\sin^2\theta + \cos^2\theta = 1\) \(\div \sin^2\theta\) \(\dfrac{\sin^2\theta}{\sin^2\theta} + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta}\) | M1 |
| \(1 + \cot^2\theta = \mathrm{cosec}^2\,\theta\ \ \ast\) cso | A1 |
| (2) |
Alternative for (a)
| \(1 + \cot^2\theta = 1 + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta}\) | M1 |
| \(= \mathrm{cosec}^2\,\theta\ \ \ast\) cso | A1 |
| Scheme | Marks |
|---|---|
| \(2(\mathrm{cosec}^2\,\theta - 1) - 9\,\mathrm{cosec}\,\theta = 3\) | M1 |
| \(2\,\mathrm{cosec}^2\,\theta - 9\,\mathrm{cosec}\,\theta - 5 = 0\) or \(5\sin^2\theta + 9\sin\theta - 2 = 0\) | M1 |
| \((2\,\mathrm{cosec}\,\theta + 1)(\mathrm{cosec}\,\theta - 5) = 0\) or \((5\sin\theta - 1)(\sin\theta + 2) = 0\) | M1 |
| \(\mathrm{cosec}\,\theta = 5\) or \(\sin\theta = \dfrac{1}{5}\) | A1 |
| \(\theta = 11.5^\circ, 168.5^\circ\) | A1 A1 |
| (6) | |
| (8 marks) |