Standard +0.3 Part (a) is a straightforward identity proof using the fundamental identity sec²x = 1 + tan²x and tan²x = sin²x/cos²x. Part (b) requires using the double angle formula cos2x = 1 - 2sin²x to form a quadratic in sin²x, then finding multiple solutions in the given range. This is a standard Further Maths Pure question combining routine algebraic manipulation with solving in an extended domain, slightly easier than average due to the direct 'hence' connection and familiar techniques.
Answer: Correct proof showing all necessary intermediate steps with no errors
Marks: 3
Guidance: M1 for either splits \(\frac{\sec^2 x - 1}{\sec^2 x} = \frac{\sec^2 x}{\sec^2 x} - \frac{1}{\sec^2 x}\) and uses \(\frac{1}{\sec^2 x} = \cos^2 x\) or states or uses \(\sec^2 x - 1 = \tan^2 x\) and \(\sec^2 x = \frac{1}{\cos^2 x}\); M1 for either replaces \(1 - \cos^2 x = \sin^2 x\) or replaces \(\frac{\tan^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} \times \cos^2 x = \frac{1}{1}\); A1 for correct proof showing all necessary intermediate steps with no errors
Guidance: M1 for uses \(\frac{\sec^2 x - 1}{\sec^2 x} = \sin^2 x\) and replaces \(\cos 2x = 1 - 2\sin^2 x\) and obtains an equation of the form \(4\sin^2 x = 1\); A1 for correct equation \(4\sin^2 x = 1\); A1 for two of \(x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°\); A1 for \(x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°\)
## Part a
**Answer:** Correct proof showing all necessary intermediate steps with no errors | **Marks:** 3 | **Guidance:** M1 for either splits $\frac{\sec^2 x - 1}{\sec^2 x} = \frac{\sec^2 x}{\sec^2 x} - \frac{1}{\sec^2 x}$ and uses $\frac{1}{\sec^2 x} = \cos^2 x$ or states or uses $\sec^2 x - 1 = \tan^2 x$ and $\sec^2 x = \frac{1}{\cos^2 x}$; M1 for either replaces $1 - \cos^2 x = \sin^2 x$ or replaces $\frac{\tan^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} \times \cos^2 x = \frac{1}{1}$; A1 for correct proof showing all necessary intermediate steps with no errors
## Part b
**Answer:** $x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°$ | **Marks:** 4 | **Guidance:** M1 for uses $\frac{\sec^2 x - 1}{\sec^2 x} = \sin^2 x$ and replaces $\cos 2x = 1 - 2\sin^2 x$ and obtains an equation of the form $4\sin^2 x = 1$; A1 for correct equation $4\sin^2 x = 1$; A1 for two of $x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°$; A1 for $x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°$
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