<s>[INST] Prove trigonometric identity: $\sin(A-D)\sin(B-D)\sin(C-D)=[\sin(D)]^3$ If $ A+B+C= 180^\circ$ and $\cot(D) =  \cot(A)+ \cot(B) + \cot(C)$
then show that 
$$\sin(A-D)\sin(B-D)\sin(C-D)=[\sin(D)]^3$$
I couldn't do much about the second given statement!!
 [/INST] \begin{align}
&\frac{\sin (A-D) \sin (B-D) \sin (C-D)}{\sin^3D} \\
&= [\sin (A) \cot (D)-\cos (A)]\cdot[\sin (B) \cot (D)-\cos (B)]\cdot[\sin (C) \cot (D)-\cos (C)]
\end{align}
Substitute  $\cot(D) =  \cot(A)+ \cot(B) + \cot(C)$ into above equation. (The algebraic detail is omitted as it's very tedious and not at all illuminating.) We can get 
\begin{align}
&\frac{\sin (A-D) \sin (B-D) \sin (C-D)}{\sin^3D} 
\\&=\frac{ \sin (A+B) \sin (A+C) \sin (B+C)}{\sin (A)\sin (B)\sin (C)}
\end{align}
Since $ A+B+C= 180^\circ$, thus $\sin(A+B)=\sin(180^\circ-C)=\sin C$, $\sin(B+C)=\sin(180^\circ-A)=\sin A$ and $\sin(A+C)=\sin(180^\circ-B)=\sin B$
Finally, we can get 
\begin{align}
&\frac{\sin (A-D) \sin (B-D) \sin (C-D)}{\sin^3D} =1
\end{align}
QED.
</s>