C3 June 2009 Q2
2.
(a) Use the identity \(\cos^2\theta + \sin^2\theta = 1\) to prove that \(\tan^2\theta = \sec^2\theta - 1\). (2)
(b) Solve, for \(0 \leqslant \theta \lt 360^\circ\), the equation\[2\tan^2\theta + 4\sec\theta + \sec^2\theta = 2\] (6)
| Scheme | Marks |
|---|---|
| \(\cos^2\theta + \sin^2\theta = 1\quad (\div \cos^2\theta)\) | |
| \(\underline{\dfrac{\cos^2\theta}{\cos^2\theta} + \dfrac{\sin^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}}\) | M1 |
| \(1 + \tan^2\theta = \sec^2\theta\) | |
| \(\tan^2\theta = \sec^2\theta - 1\) (as required) AG | A1 cso |
| (2) |
Notes
M1: Dividing \(\cos^2\theta + \sin^2\theta = 1\) by \(\cos^2\theta\) to give underlined equation.
A1 cso: Complete proof. No errors seen.
| Scheme | Marks |
|---|---|
| \(2\tan^2\theta + 4\sec\theta + \sec^2\theta = 2\), (eqn \(*\)) \(\quad 0 \leqslant \theta \lt 360^\circ\) | |
| \(2(\sec^2\theta - 1) + 4\sec\theta + \sec^2\theta = 2\) | M1 |
| \(2\sec^2\theta - 2 + 4\sec\theta + \sec^2\theta = 2\) | |
| \(\underline{3\sec^2\theta + 4\sec\theta - 4 = 0}\) | M1 |
| \((\sec\theta + 2)(3\sec\theta - 2) = 0\) | M1 |
| \(\sec\theta = -2\) or \(\sec\theta = \tfrac{2}{3}\) \(\dfrac{1}{\cos\theta} = -2\) or \(\dfrac{1}{\cos\theta} = \dfrac{2}{3}\) | |
| \(\underline{\cos\theta = -\tfrac{1}{2}}\); or \(\cos\theta = \tfrac{3}{2}\) | A1; |
| \(\alpha = 120^\circ\) or \(\alpha = \) no solutions | |
| \(\theta_1 = \underline{120^\circ}\) | A1 |
| \(\theta_2 = 240^\circ\) | B1ft |
| \(\theta = \{120^\circ, 240^\circ\}\) | |
| (6) | |
| (8 marks) |
Notes
1st M1: Substituting \(\tan^2\theta = \sec^2\theta - 1\) into eqn \(*\) to get a quadratic in \(\sec\theta\) only
2nd M1: Forming a three term “one sided” quadratic expression in \(\sec\theta\).
3rd M1: Attempt to factorise or solve a quadratic.
A1: \(\underline{\cos\theta = -\tfrac{1}{2}}\)
A1: \(\underline{120^\circ}\)
B1ft: \(\underline{240^\circ}\) or \(\theta_2 = 360^\circ - \theta_1\) when solving using \(\cos\theta = \ldots\)
Note the final A1 mark has been changed to a B1 mark.