Answer: \(a=16\), \(b=-5\), \(c=10\).

Hence

\(\operatorname{cosec}^5\theta=\dfrac{16}{\sin 5\theta-5\sin 3\theta+10\sin\theta}.\)

Given \(z=\cos\theta+\mathrm{i}\sin\theta\), we have

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

so

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

Now expand and group:

\((z-z^{-1})^5=z^5-5z^3+10z-10z^{-1}+5z^{-3}-z^{-5}\)

so

\((z-z^{-1})^5=(z^5-z^{-5})-5(z^3-z^{-3})+10(z-z^{-1}).\)

Using de Moivre's theorem,

\(z^n=\cos n\theta+\mathrm{i}\sin n\theta\), \(z^{-n}=\cos n\theta-\mathrm{i}\sin n\theta\),

therefore

\(z^n-z^{-n}=2\mathrm{i}\sin n\theta.\)

Hence

\((z-z^{-1})^5=2\mathrm{i}\sin5\theta-5(2\mathrm{i}\sin3\theta)+10(2\mathrm{i}\sin\theta).\)

But also

\((z-z^{-1})^5=(2\mathrm{i}\sin\theta)^5=2^5\mathrm{i}^5\sin^5\theta=32\mathrm{i}\sin^5\theta.\)

So

\(32\mathrm{i}\sin^5\theta=2\mathrm{i}(\sin5\theta-5\sin3\theta+10\sin\theta).\)

Dividing by \(2\mathrm{i}\),

\(16\sin^5\theta=\sin5\theta-5\sin3\theta+10\sin\theta.\)

Therefore

\(\operatorname{cosec}^5\theta=\dfrac{1}{\sin^5\theta}=\dfrac{16}{\sin5\theta-5\sin3\theta+10\sin\theta}.\)

Thus \(a=16\), \(b=-5\), \(c=10\).