Answer: \(\tan5\theta=\frac{5t-10t^3+t^5}{1-10t^2+5t^4},\qquad t=\tan\theta.\)
The roots of \(t^4-10t^2+5=0\) are
\(\pm\tan\frac{\pi}{5},\qquad \pm\tan\frac{2\pi}{5}.\)
Also,
\(\tan\frac{\pi}{5}\tan\frac{2\pi}{5}=\sqrt5.\)
Let \(t=\tan\theta\). From de Moivre’s theorem,
\(\,(\cos\theta+i\sin\theta)^5=\cos 5\theta+i\sin 5\theta\,\).
Expanding the left-hand side gives
\(\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta\).
Equating real and imaginary parts:
\(\cos 5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\),
\(\sin 5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta\).
Now factor out \(\cos^5\theta\) in each expression:
\(\cos 5\theta=\cos^5\theta\bigl(1-10\tan^2\theta+5\tan^4\theta\bigr)\),
\(\sin 5\theta=\cos^5\theta\bigl(5\tan\theta-10\tan^3\theta+\tan^5\theta\bigr)\).
Since \(t=\tan\theta\), this gives
\(\tan 5\theta=\dfrac{\sin 5\theta}{\cos 5\theta}=\dfrac{5t-10t^3+t^5}{1-10t^2+5t^4}\),
which is the required identity.
Next, let \(\theta=\frac{\pi}{5}\) and \(\theta=\frac{2\pi}{5}\). Then \(5\theta=\pi\) and \(5\theta=2\pi\), so \(\tan 5\theta=0\). From
\(\tan 5\theta=\dfrac{5t-10t^3+t^5}{1-10t^2+5t^4}\),
the numerator must be zero:
\(5t-10t^3+t^5=0\).
Factorising,
\(t(t^4-10t^2+5)=0\).
Thus the non-zero values of \(t\) satisfy
\(t^4-10t^2+5=0\).
Therefore the roots are \(t=\pm\tan\frac{\pi}{5}\) and \(t=\pm\tan\frac{2\pi}{5}\).
Finally, for \(t^4-10t^2+5=0\), let \(x=t^2\). Then
\(x^2-10x+5=0\).
The product of the roots of this quadratic is \(5\), so
\(\tan^2\frac{\pi}{5}\tan^2\frac{2\pi}{5}=5\).
Since both tangents are positive,
\(\tan\frac{\pi}{5}\tan\frac{2\pi}{5}=\sqrt{5}\).
