Standard +0.3 This is a straightforward application of the Pythagorean identity sec²θ = 1 + tan²θ to convert to a quadratic in tan θ, then solve. It requires standard technique with no novel insight, making it slightly easier than average for A-level.
solving their quadratic for \(\tan\theta\) (follow rules for solving as in Question 1)
A1
www
A1
first correct solution (or better)
A1
second correct solution (or better) and no others in the range. Ignore solutions outside the range. SC A1 for both 0.46 and 1.11. SC A1 for both 26.6° and 63.4° (or better). Do not award SCs if there are extra solutions in range.
Solution 2:
Answer
Marks
M1
Using \(\sec\theta = \frac{1}{\cos\theta}\) and \(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
A1
\(2 - 5\sin\theta\cos\theta = 0\) or \(2\cos\theta = 5\sin\theta\cos^2\theta\) oe (or common denominator). Do not need \(\cos\theta \neq 0\) at this stage.
M1
Using \(\sin 2\theta = 2\sin\theta\cos\theta\) oe, e.g. \(2 = 5\sin\theta\sqrt{1-\sin^2\theta}\) and squaring
second correct solution (or better) and no others in the range. Ignore solutions outside the range. SCs as above.
# Question 1
**Solution 1:**
M1 | $\sec^2\theta = 1 + \tan^2\theta$ used
A1 | $2(1 + \tan^2\theta) = 5\tan\theta$
M1 | $2\tan^2\theta - 5\tan\theta + 2 = 0$
A1 | $(2\tan\theta - 1)(\tan\theta - 2) = 0$
A1 | $\tan\theta = \frac{1}{2}$ or $2$
A1 | $\theta = 0.464, 1.107$ (radians)
**Guidance for Solution 1:**
M1 | $\sec^2\theta = 1 + \tan^2\theta$ used
A1 | correct quadratic oe
M1 | solving their quadratic for $\tan\theta$ (follow rules for solving as in Question 1)
A1 | www
A1 | first correct solution (or better)
A1 | second correct solution (or better) and no others in the range. Ignore solutions outside the range. SC A1 for both 0.46 and 1.11. SC A1 for both 26.6° and 63.4° (or better). Do not award SCs if there are extra solutions in range.
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**Solution 2:**
M1 | Using $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$
A1 | $2 - 5\sin\theta\cos\theta = 0$ or $2\cos\theta = 5\sin\theta\cos^2\theta$ oe (or common denominator). Do not need $\cos\theta \neq 0$ at this stage.
M1 | Using $\sin 2\theta = 2\sin\theta\cos\theta$ oe, e.g. $2 = 5\sin\theta\sqrt{1-\sin^2\theta}$ and squaring
A1 | $\sin 2\theta = 0.8$ or, say, $25\sin^4\theta - 25\sin^2\theta + 4 = 0$
A1 | first correct solution (or better)
A1 | second correct solution (or better) and no others in the range. Ignore solutions outside the range. SCs as above.