10 Using de Moivre's theorem, show that
$$\tan 5 \theta = \frac { 5 \tan \theta - 10 \tan ^ { 3 } \theta + \tan ^ { 5 } \theta } { 1 - 10 \tan ^ { 2 } \theta + 5 \tan ^ { 4 } \theta }$$
Hence show that the equation \(x ^ { 2 } - 10 x + 5 = 0\) has roots \(\tan ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)\) and \(\tan ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)\).
Deduce a quadratic equation, with integer coefficients, having roots \(\sec ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)\) and \(\sec ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)\).
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10 Using de Moivre's theorem, show that
$$\tan 5 \theta = \frac { 5 \tan \theta - 10 \tan ^ { 3 } \theta + \tan ^ { 5 } \theta } { 1 - 10 \tan ^ { 2 } \theta + 5 \tan ^ { 4 } \theta }$$
Hence show that the equation $x ^ { 2 } - 10 x + 5 = 0$ has roots $\tan ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)$ and $\tan ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)$.
Deduce a quadratic equation, with integer coefficients, having roots $\sec ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)$ and $\sec ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)$.\\[0pt]
[Question 11 is printed on the next page.]
\hfill \mbox{\textit{CAIE FP1 2015 Q10}}