Prove trigonometric identity: $\sin(A-D)\sin(B-D)\sin(C-D)=[\sin(D)]^3$

Question

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!!

Answer 1

HINT:

$$\cot D-\cot A=\cot B+\cot C$$

$$\implies\frac{\sin(A-D)}{\sin D\sin A}=\frac{\sin(B+C)}{\sin B\sin C}$$

Now $\sin(B+C)=\sin(\pi-A)=\sin A$

$$\implies\sin(A-D)=?$$

Answer 2

\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.