C3 June 2005 Q1
1.
(a) Given that \(\sin^2\theta + \cos^2\theta \equiv 1\), show that \(1 + \tan^2\theta \equiv \sec^2\theta\). (2)
(b) Solve, for \(0 \leqslant \theta \lt 360^\circ\), the equation\[2\tan^2\theta + \sec\theta = 1,\]giving your answers to 1 decimal place. (6)
| Scheme | Marks |
|---|---|
| Dividing by \(\cos^2\theta\): \(\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} \equiv \dfrac{1}{\cos^2\theta}\) | M1 |
| Completion: \(1 + \tan^2\theta \equiv \sec^2\theta\) (no errors seen) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Use of \(1 + \tan^2\theta = \sec^2\theta\): \(2(\sec^2\theta - 1) + \sec\theta = 1\) \([\,2\sec^2\theta + \sec\theta - 3 = 0\,]\) | M1 |
| Factorising or solving: \((2\sec\theta + 3)(\sec\theta - 1) = 0\) \([\,\sec\theta = -\tfrac{3}{2}\) or \(\sec\theta = 1\,]\) | M1 |
| \(\theta = 0\) | B1 |
| \(\cos\theta = -\tfrac{2}{3}\); \(\quad\theta_1 = 131.8^\circ\) | M1 A1 |
| \(\theta_2 = 228.2^\circ\) | A1ft |
| (6) | |
| (8 marks) |
Notes
[A1ft for \(\theta_2 = 360^\circ - \theta_1\)]