## Question 2:

### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use $2\cos^2\alpha - 1$ or $\cos^2\alpha - \sin^2\alpha$ or $1-2\sin^2\alpha$ | B1 | |
| Obtain equation in which $\sin^2\alpha$ appears once | M1 | Condoning sign slips or arithmetic slips; for solution which gives equation involving $\tan^2\alpha$, M1 is not earned until valid method for reaching $\sin\alpha$ is used; attempt involving $4(1-s^2)=s^2$ is M0 |
| Obtain $\pm\frac{2}{3}$ | A1 | Both values needed; $\pm 0.667$ is A0; $\pm\sqrt{\frac{4}{9}}$ is A0; ignore subsequent work to find angle(s) |

### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either:** Attempt use of identity | M1 | Of form $\tan^2\beta = \pm\sec^2\beta \pm 1$ |
| Obtain $2\sec^2\beta - 9\sec\beta - 5 = 0$ | A1 | Condone absence of $= 0$ |
| Attempt solution of 3-term quadratic in $\sec\beta$ to obtain at least one value of $\sec\beta$ | M1 | If factorising, factors must be such that expansion gives their first and third terms; if using formula, this must be correct for their values |
| Obtain 5 with no errors in solution | A1 | And, finally, no other value; no need to justify rejection of $-\frac{1}{2}$ |
| **Or:** Attempt to express equation in terms of $\cos\beta$ | M1 | Using identities which are correct apart maybe for sign slips |
| Obtain $5\cos^2\beta + 9\cos\beta - 2 = 0$ | A1 | Condone absence of $= 0$ |
| Attempt solution of 3-term quadratic and show switch at least once to a secant value | M1 | If factorising, factors must be such that expansion gives their first and third terms |
| Obtain 5 with no errors in solution | A1 | And, finally, no other value; no need to justify rejection of $-\frac{1}{2}$ |

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