Answer:

Using de Moivre's theorem,

\((\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta.\)

Expanding and equating imaginary and real parts gives

\(\sin5\theta=5\cos^{4}\theta\sin\theta-10\cos^{2}\theta\sin^{3}\theta+\sin^{5}\theta,\)

and

\(\cos5\theta=\cos^{5}\theta-10\cos^{3}\theta\sin^{2}\theta+5\cos\theta\sin^{4}\theta.\)

Divide numerator and denominator of \(\tan5\theta=\frac{\sin5\theta}{\cos5\theta}\) by \(\cos^{5}\theta\):

\(\tan 5\theta=\frac{5\tan\theta-10\tan^{3}\theta+\tan^{5}\theta}{1-10\tan^{2}\theta+5\tan^{4}\theta}.\)

For \(\theta=\frac{\pi}{5}\) and \(\theta=\frac{2\pi}{5}\), \(5\theta\) is an integer multiple of \(\pi\), so \(\tan5\theta=0\). Since \(\tan\theta\neq0\), the numerator gives

\(5-10\tan^{2}\theta+\tan^{4}\theta=0.\)

Thus, if \(x=\tan^{2}\theta\),

\(x^{2}-10x+5=0.\)

So the roots of this equation are \(\tan^{2}\frac{\pi}{5}\) and \(\tan^{2}\frac{2\pi}{5}\).

Now

\(\sec^{2}\theta=1+\tan^{2}\theta.\)

If \(y=\sec^{2}\theta\), then \(x=y-1\). Substitute \(x=y-1\) into \(x^{2}-10x+5=0\):

\((y-1)^{2}-10(y-1)+5=0.\)

This simplifies to

\(y^{2}-12y+16=0.\)