91Ó°ÊÓ

The given equations are

$$\begin{gathered} \mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}={180}^{∘} \\ cot\mathrm{θ}=cot\mathrm{Î\pm }+cot\mathrm{β}+cot\mathrm{γ} \\ 0<\mathrm{θ}<{180}^{∘} \end{gathered}$$

The equation we need to prove is

${\sin }^3\mathrm{θ}=\sin (\mathrm{Î\pm }-\mathrm{θ})\sin (\mathrm{β}-\mathrm{θ})\sin (\mathrm{γ}-\mathrm{θ})$

As $cot\mathrm{θ}=cot\mathrm{Î\pm }+cot\mathrm{β}+cot\mathrm{γ}$

So

$$\begin{gathered} cot\mathrm{θ}=cot\mathrm{Î\pm }+cot\mathrm{β}+cot\mathrm{γ} \\ cot\mathrm{θ}-cot\mathrm{Î\pm }=cot\mathrm{β}+cot\mathrm{γ} \\ \frac{\cos \mathrm{θ}}{\sin \mathrm{θ}}-\frac{\cos \mathrm{Î\pm }}{\sin \mathrm{Î\pm }}=\frac{\cos \mathrm{β}}{\sin \mathrm{β}}+\frac{\cos \mathrm{γ}}{\sin \mathrm{γ}} \\ \frac{\cos \mathrm{θ}\sin \mathrm{Î\pm }-\cos \mathrm{Î\pm }\sin \mathrm{θ}}{\sin \mathrm{θ}\sin \mathrm{Î\pm }}=\frac{\cos \mathrm{β}\sin \mathrm{γ}+\cos \mathrm{γ}\sin \mathrm{β}}{\sin \mathrm{β}\sin \mathrm{γ}} \\ \frac{\sin (\mathrm{Î\pm }-\mathrm{θ})}{\sin \mathrm{θ}\sin \mathrm{Î\pm }}=\frac{\sin (\mathrm{γ}+\mathrm{β})}{\sin \mathrm{β}\sin \mathrm{γ}} \\ and\mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}={180}^{∘} \\ \mathrm{β}+\mathrm{γ}={180}^{∘}-\mathrm{Î\pm } \\ So\frac{\sin (\mathrm{Î\pm }-\mathrm{θ})}{\sin \mathrm{θ}\sin \mathrm{Î\pm }}=\frac{\sin ({180}^{∘}-\mathrm{Î\pm })}{\sin \mathrm{β}\sin \mathrm{γ}} \\ \frac{\sin (\mathrm{Î\pm }-\mathrm{θ})}{\sin \mathrm{θ}\sin \mathrm{Î\pm }}=\frac{\sin \left(\mathrm{Î\pm }\right)}{\sin \mathrm{β}\sin \mathrm{γ}} \\ \sin (\mathrm{Î\pm }-\mathrm{θ})=\frac{{\sin }^2\mathrm{Î\pm }\sin \mathrm{θ}}{\sin \mathrm{β}\sin \mathrm{γ}}\left(i\right) \end{gathered}$$

Similarly

$$\begin{gathered} \sin (\mathrm{β}-\mathrm{θ})=\frac{{\sin }^2\mathrm{β}\sin \mathrm{θ}}{\sin \mathrm{Î\pm }\sin \mathrm{γ}}\left(ii\right) \\ \sin (\mathrm{γ}-\mathrm{θ})=\frac{{\sin }^2\mathrm{γ}\sin \mathrm{θ}}{\sin \mathrm{β}\sin \mathrm{Î\pm }}\left(iii\right) \end{gathered}$$

Multiply equation(i),(ii),and(iii)

$$\begin{gathered} \sin (\mathrm{Î\pm }-\mathrm{θ})\sin (\mathrm{β}-\mathrm{θ})\sin (\mathrm{γ}-\mathrm{θ})=\frac{{\sin }^2\mathrm{Î\pm }\sin \mathrm{θ}}{\sin \mathrm{β}\sin \mathrm{γ}}×\frac{{\sin }^2\mathrm{β}\sin \mathrm{θ}}{\sin \mathrm{Î\pm }\sin \mathrm{γ}}×\frac{{\sin }^2\mathrm{γ}\sin \mathrm{θ}}{\sin \mathrm{β}\sin \mathrm{Î\pm }} \\ \sin (\mathrm{Î\pm }-\mathrm{θ})\sin (\mathrm{β}-\mathrm{θ})\sin (\mathrm{γ}-\mathrm{θ})={\sin }^2\mathrm{θ} \end{gathered}$$