Standard +0.8 This question requires applying the tan addition formula, converting cot to tan, then solving a resulting quadratic in tan(x). It involves multiple trigonometric manipulations and algebraic steps beyond routine exercises, but follows a standard problem-solving pattern for this topic without requiring exceptional insight.
Use correct \(\tan(A+B)\) formula and obtain an equation in \(\tan x\) or an equation in \(\cos x\) and \(\sin x\)
M1
e.g. \(\frac{\tan x + \tan 45°}{1 - \tan x \tan 45°} = \frac{2}{\tan x}\); or \(\frac{\sin x \cos 45° + \cos x \sin 45°}{\cos x \cos 45° - \sin x \sin 45°} = \frac{2\cos x}{\sin x}\)
Obtain correct 3-term equation \(\tan^2 x + 3\tan x - 2 = 0\), or equivalent
A1
or \(3\sin x \cos x = 2\cos^2 x - \sin^2 x\)
Solve a 3-term quadratic in \(\tan x\) and obtain a value for \(x\)
M1
Obtain answer, e.g. \(29.3°\)
A1
\(29.316\ldots\)
Obtain second answer, e.g. \(105.7°\) and no other
A1
\(105.583\ldots\); ignore answers outside the given interval; treat answers in radians as a misread
Total
5
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct $\tan(A+B)$ formula and obtain an equation in $\tan x$ or an equation in $\cos x$ and $\sin x$ | M1 | e.g. $\frac{\tan x + \tan 45°}{1 - \tan x \tan 45°} = \frac{2}{\tan x}$; or $\frac{\sin x \cos 45° + \cos x \sin 45°}{\cos x \cos 45° - \sin x \sin 45°} = \frac{2\cos x}{\sin x}$ |
| Obtain correct 3-term equation $\tan^2 x + 3\tan x - 2 = 0$, or equivalent | A1 | or $3\sin x \cos x = 2\cos^2 x - \sin^2 x$ |
| Solve a 3-term quadratic in $\tan x$ and obtain a value for $x$ | M1 | |
| Obtain answer, e.g. $29.3°$ | A1 | $29.316\ldots$ |
| Obtain second answer, e.g. $105.7°$ and no other | A1 | $105.583\ldots$; ignore answers outside the given interval; treat answers in radians as a misread |
| **Total** | **5** | |
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