Moderate -0.3 Part (a) requires recognizing the Pythagorean identity (1 - sin²x = cos²x) and simplifying to reach 1 - tan²x, which is straightforward algebraic manipulation. Part (b) is a direct application of the proven identity leading to tan²x = 3, then solving a basic quadratic-type trig equation. This is slightly easier than average as it's a standard two-part question with clear scaffolding and routine techniques.
Completes proof with no errors. Show that question — look for minimum of ALL 3 steps: \(\frac{\cos^2 x - \sin^2 x}{\cos^2 x} = 1 - \frac{\sin^2 x}{\cos^2 x} = 1-\tan^2 x\)
Part (b):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
\(\frac{\cos^2 x - \sin^2 x}{1-\sin^2 x} + 2 = 0 \Rightarrow 1 - \tan^2 x + 2 = 0\)
M1
Scored for using part (a) result to reach \(\tan^2 x = k\)
\(\tan^2 x = 3\)
A1
\(\tan x = (\pm)\sqrt{3} \Rightarrow x = \ldots\)
dM1
Correct order: \(\tan^2 x = k \Rightarrow \tan x = (\pm)\sqrt{k}\), no need for negative, leading to at least one value of \(x\). If \(k < 0\) this mark cannot be scored
First A1 for two correct answers as multiples of \(\pi\). Second A1 for all four correct with no extras inside range. Ignore extra solutions outside range. Degrees/decimals score A0 although recovery allowed
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $1 - \sin^2 x = \cos^2 x$ | M1 | Used in denominator. Usual case: $\frac{\cos^2 x - \sin^2 x}{1-\sin^2 x} \rightarrow \frac{\cos^2 x - \sin^2 x}{\cos^2 x}$ |
| $\frac{\cos^2 x - \sin^2 x}{1-\sin^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x} = 1 - \frac{\sin^2 x}{\cos^2 x} = 1 - \tan^2 x$ | A1* | Completes proof with no errors. Show that question — look for minimum of ALL 3 steps: $\frac{\cos^2 x - \sin^2 x}{\cos^2 x} = 1 - \frac{\sin^2 x}{\cos^2 x} = 1-\tan^2 x$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\cos^2 x - \sin^2 x}{1-\sin^2 x} + 2 = 0 \Rightarrow 1 - \tan^2 x + 2 = 0$ | M1 | Scored for using part (a) result to reach $\tan^2 x = k$ |
| $\tan^2 x = 3$ | A1 | |
| $\tan x = (\pm)\sqrt{3} \Rightarrow x = \ldots$ | dM1 | Correct order: $\tan^2 x = k \Rightarrow \tan x = (\pm)\sqrt{k}$, no need for negative, leading to at least one value of $x$. If $k < 0$ this mark cannot be scored |
| $x = \frac{1}{3}\pi, \frac{2}{3}\pi, \frac{4}{3}\pi, \frac{5}{3}\pi$ | A1, A1 | First A1 for two correct answers as multiples of $\pi$. Second A1 for all four correct with no extras inside range. Ignore extra solutions outside range. Degrees/decimals score A0 although recovery allowed |
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