Standard +0.3 This question requires converting reciprocal trig functions to standard form, applying the Pythagorean identity, and solving a quadratic in sin θ. While it involves multiple steps and reciprocal functions, the technique is standard for P2 level with no novel insight required—slightly easier than average due to straightforward algebraic manipulation once the identity is applied.
State \(\frac{1}{\cos^2\theta} = \frac{3}{\sin\theta}\) or \(1 + \tan^2\theta = \frac{3}{\sin\theta}\)
B1
Produce quadratic equation in \(\sin\theta\)
M1
Dependent on B1
Solve 3-term quadratic equation to find value between \(-1\) and \(1\) for \(\sin\theta\)
M1
Dependent on first M1
Obtain \(\sin\theta = \frac{1}{6}(-1+\sqrt{37})\) and hence 57.9
A1
Obtain 122.1 and no others between 0 and 180
A1
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $\frac{1}{\cos^2\theta} = \frac{3}{\sin\theta}$ or $1 + \tan^2\theta = \frac{3}{\sin\theta}$ | B1 | |
| Produce quadratic equation in $\sin\theta$ | M1 | Dependent on B1 |
| Solve 3-term quadratic equation to find value between $-1$ and $1$ for $\sin\theta$ | M1 | Dependent on first M1 |
| Obtain $\sin\theta = \frac{1}{6}(-1+\sqrt{37})$ and hence 57.9 | A1 | |
| Obtain 122.1 and no others between 0 and 180 | A1 | |
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