**Answer:** $\operatorname{cosec}^2\theta = 1 + \cot^2\theta$

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

$\Rightarrow \cot^2\theta - 2\cot\theta = 0$

$\Rightarrow \cot\theta(\cot\theta - 2) = 0$

$\Rightarrow \cot\theta = 0,$ **and** $\cot\theta = 2, \tan\theta = \frac{1}{2}$

$\theta = 26.6°, -153.4°, -90°, 90°$ 
...................................................................

**OR** $\frac{1}{\sin^2\theta} = 1 + \frac{2\cos\theta}{\sin\theta} + \frac{2\cos\theta}{\sin\theta}$

$\Rightarrow \sin^2\theta + 2\sin\theta\cos\theta - 1 = 0$

$\Rightarrow 2\sin\theta\cos\theta - \cos^2\theta = 0$

$\Rightarrow \cos\theta(\sin\theta - \cos\theta) = 0$

$\Rightarrow \cos\theta = 0,$ **and** $\tan\theta = \frac{1}{2}$

$\theta = 26.6°, -153.4°, -90°, 90°$

| M1 | correct trig identity used |
| M1 | factorising oe |
| M1 | both needed and $\cot\theta = 1/\tan\theta$ soi |
| B3,2,1,0 | accept 27°, -153° as above |
| | (omission of $\cot\theta=0$ could gain M1, M1, M0, B1) ........................................... |
| M1 | correct trig equivalents and a one line equation (or common denominator) formed |
| M1 | use of Pythagoras and factorising |
| M1 | accept 27°, -153° as above |
| B3,2,1,0 | |
| | **in both cases, -1 if extra solutions in the range are given ( dependent on at least B1 being scored)-not their incorrect solutions eg 26.6°,-153.4°, 0°, 180°, -180° would obtain B1 to lose both of these, at least B2 would need to be scored.** |

**Total: [6]**

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