Standard +0.8 This question requires converting cotangent to tangent, applying the compound angle formula for tan(θ + 45°), algebraic manipulation to form a quadratic, and solving within a specified range. It combines multiple techniques (reciprocal trig identities, addition formulae, quadratic solving) in a non-routine way that goes beyond standard textbook exercises, making it moderately challenging but still within typical A-level scope.
Use correct trig formula and obtain an equation in \(\tan\theta\)
M1
Allow with 45°: \(\frac{1}{\tan\theta} - \frac{1}{\frac{\tan\theta+\tan 45°}{1-\tan\theta\tan 45°}} = 3\)
Obtain a correct horizontal equation in any form
A1
e.g. \(1 + \tan\theta - \tan\theta(1-\tan\theta) = 3\tan\theta(1+\tan\theta)\)
Reduce to \(2\tan^2\theta + 3\tan\theta - 1 = 0\)
A1
or 3-term equivalent
Solve 3-term quadratic and find a value of \(\theta\)
M1
Must see working if using an incorrect quadratic
Obtain answer \(15.7°\)
A1
One correct solution (degrees to at least 3 sf)
Obtain answer \(119.3°\)
A1
Second correct solution and no others in range (degrees to at least 3 sf). Mark 0.274, 2.082 as MR: A0A1
Total
6
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct trig formula and obtain an equation in $\tan\theta$ | M1 | Allow with 45°: $\frac{1}{\tan\theta} - \frac{1}{\frac{\tan\theta+\tan 45°}{1-\tan\theta\tan 45°}} = 3$ |
| Obtain a correct horizontal equation in any form | A1 | e.g. $1 + \tan\theta - \tan\theta(1-\tan\theta) = 3\tan\theta(1+\tan\theta)$ |
| Reduce to $2\tan^2\theta + 3\tan\theta - 1 = 0$ | A1 | or 3-term equivalent |
| Solve 3-term quadratic and find a value of $\theta$ | M1 | Must see working if using an incorrect quadratic |
| Obtain answer $15.7°$ | A1 | One correct solution (degrees to at least 3 sf) |
| Obtain answer $119.3°$ | A1 | Second correct solution and no others in range (degrees to at least 3 sf). Mark 0.274, 2.082 as MR: A0A1 |
| **Total** | **6** | |