Standard +0.8 This question requires proving equivalence of two expressions involving complex number arguments and tangent identities, likely requiring manipulation of inverse tangent functions and understanding of argument properties. It's above average difficulty as it demands both technical facility with trigonometric identities and careful reasoning about equivalence, but is still within standard A-level pure mathematics scope.
\(\frac{\sqrt{6}-\sqrt{2}}{2}\) is positive so it is equal to \(\sqrt{2-\sqrt{3}}\)
E1
Completion of argument to show the two values are equal
# Question 14:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)^2 = \frac{8-2\sqrt{12}}{4}$ | M1 | Attempt to square |
| $= \frac{8-4\sqrt{3}}{4} = 2-\sqrt{3}$ | A1 | Answer in exact form |
| $\frac{\sqrt{6}-\sqrt{2}}{2}$ is positive so it is equal to $\sqrt{2-\sqrt{3}}$ | E1 | Completion of argument to show the two values are equal |
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