## Question 5:

$\cosec^2\theta = 1 + \cot^2\theta$

| Answer | Mark | Guidance |
|--------|------|----------|
| $1 + \cot^2\theta = 1 + 2\cot\theta$ | M1 | Correct trig identity used. (Use of $1 - \cot^2\theta$ could lead to M0 M1 M1 B1) |
| $\cot^2\theta - 2\cot\theta = 0$ | | |
| $\cot\theta(\cot\theta - 2) = 0$ | M1 | Factorising oe. Allow if $\cot\theta = 0$ not seen (ie quadratic equation followed by $\cot\theta - 2 = 0$ or $\cot\theta = 2$) |
| $\cot\theta = 0$, **and** $\cot\theta = 2$, $\tan\theta = \frac{1}{2}$ | M1 | **Both** needed and $\cot\theta = 1/\tan\theta$ soi |
| $\Rightarrow \theta = 26.6°, -153.4°, -90°, 90°$ | B3,2,1,0 | $-90°, 90°, 27°, -153°$ or better www. (Omission of $\cot\theta = 0$ could gain M1, M1, M0, B1) |
| **OR** $\frac{1}{\sin^2\theta} = 1 + \frac{2\cos\theta}{\sin\theta} = \frac{\sin\theta + 2\cos\theta}{\sin\theta}$ | M1 | Correct trig equivalents and a one line equation (or common denominator) formed |
| $\Rightarrow \sin^2\theta + 2\sin\theta\cos\theta - 1 = 0$ | | |
| $\Rightarrow 2\sin\theta\cos\theta - \cos^2\theta = 0$ | M1 | Use of Pythagoras and factorising. As above. Allow if $\cos\theta = 0$ not seen |
| $\Rightarrow \cos\theta(2\sin\theta - \cos\theta) = 0$ | M1 | **Both** needed and $\tan\theta = \sin\theta/\cos\theta$ oe soi |
| $\Rightarrow \cos\theta = 0$, **and** $\tan\theta = \frac{1}{2}$ | | |
| $\theta = 26.6°, -153.4°, -90°, 90°$ | B3,2,1,0 | Accept $27°, -153°$ as above |
| | **[6]** | In both cases: -1 if **extra** solutions in range given (dependent on at least B1 being scored). $26.6°, -153.4°, 0°, 180°, -180°$ would obtain B1. -1 MR if answers given in radians ($-\pi/2, \pi/2, 0.464, -2.68$ ($-1.57, 1.57$) or multiples of $\pi$ that round to these, or better) (dependent on at least B1 being scored). To lose both, at least B2 would need to be scored. |

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