 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/5/a/solution

1 = 2 t / (1 - t^{2})

. Here

0 < t < 1

, because

0 < π /8 < π /4

, so the denominator is nonzero. Rearranging yields the quadratic equation

t^{2} + 2 t - 1 = 0.

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