Answer: \(a=4\) and \(b=3\), so \(\cot^{4}\theta=\frac{\cos 4\theta+4\cos 2\theta+3}{\cos 4\theta-4\cos 2\theta+3}\).

Let \(z=\cos\theta+\mathrm{i}\sin\theta\). Then \(z^{-1}=\cos\theta-\mathrm{i}\sin\theta\), so

\(z+z^{-1}=2\cos\theta\), \quad \(z-z^{-1}=2\mathrm{i}\sin\theta\).

Now expand:

\(\left(z+\frac1z\right)^4=z^4+4z^2+6+4z^{-2}+z^{-4}=\left(z^4+z^{-4}\right)+4\left(z^2+z^{-2}\right)+6.\)

Also,

\(\left(z-\frac1z\right)^4=z^4-4z^2+6-4z^{-2}+z^{-4}=\left(z^4+z^{-4}\right)-4\left(z^2+z^{-2}\right)+6.\)

Using de Moivre's theorem, \(z^n+z^{-n}=2\cos n\theta\). Hence

\(\left(z+z^{-1}\right)^4=2\cos4\theta+4(2\cos2\theta)+6=2\cos4\theta+8\cos2\theta+6,\)

\(\left(z-z^{-1}\right)^4=2\cos4\theta-8\cos2\theta+6.\)

But from \(z+z^{-1}=2\cos\theta\) and \(z-z^{-1}=2\mathrm{i}\sin\theta\),

\(\left(z+z^{-1}\right)^4=(2\cos\theta)^4=16\cos^4\theta,\)

\(\left(z-z^{-1}\right)^4=(2\mathrm{i}\sin\theta)^4=16\sin^4\theta.\)

Therefore

\(\frac{16\cos^4\theta}{16\sin^4\theta}=\frac{2\cos4\theta+8\cos2\theta+6}{2\cos4\theta-8\cos2\theta+6}.\)

So

\(\cot^4\theta=\frac{2\cos4\theta+8\cos2\theta+6}{2\cos4\theta-8\cos2\theta+6}=\frac{\cos4\theta+4\cos2\theta+3}{\cos4\theta-4\cos2\theta+3}.\)

Hence \(a=4\) and \(b=3\).