If $ \alpha +\beta +\gamma =\pi $ , then $ \cos \alpha \sin \left( \beta -\gamma \right)+\cos \beta \sin \left( \gamma -\alpha \right)+\cos \gamma \sin \left( \alpha -\beta \right) $
A. 0
B. $ \dfrac{1}{2} $
C. 1
D. $ 4\cos \alpha \cos \beta \cos \gamma $

Hint: We first make the changes for the angles where $ \alpha =\pi -\left( \beta +\gamma \right) $ , $ \beta =\pi -\left( \alpha +\gamma \right) $ , $ \gamma =\pi -\left( \alpha +\beta \right) $ . We take the cos ratio for all three angles. We replace those values in the main equation. We use the sum of angles theorem $ 2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right) $ . We get the solution.

Complete step-by-step answer:
The given condition is $ \alpha +\beta +\gamma =\pi $ . We get $ \alpha =\pi -\left( \beta +\gamma \right) $ , $ \beta =\pi -\left( \alpha +\gamma \right) $ , $ \gamma =\pi -\left( \alpha +\beta \right) $
Applying ratio cos on three conditions, we get
 $ \cos \alpha =\cos \left[ \pi -\left( \beta +\gamma \right) \right]=-\cos \left( \beta +\gamma \right) $
 $ \cos \beta =\cos \left[ \pi -\left( \alpha +\gamma \right) \right]=-\cos \left( \alpha +\gamma \right) $
 $ \cos \gamma =\cos \left[ \pi -\left( \alpha +\beta \right) \right]=-\cos \left( \alpha +\beta \right) $
The main equation becomes
\[\begin{align}
  & \cos \alpha \sin \left( \beta -\gamma \right)+\cos \beta \sin \left( \gamma -\alpha \right)+\cos \gamma \sin \left( \alpha -\beta \right) \\
 & =-\cos \left( \beta +\gamma \right)\sin \left( \beta -\gamma \right)-\cos \left( \alpha +\gamma \right)\sin \left( \gamma -\alpha \right)-\cos \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right) \\
 & =-\dfrac{1}{2}\left[ 2\cos \left( \beta +\gamma \right)\sin \left( \beta -\gamma \right)+2\cos \left( \alpha +\gamma \right)\sin \left( \gamma -\alpha \right)+2\cos \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right) \right] \\
\end{align}\]
We use the theorem $ 2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right) $ .
For \[2\cos \left( \beta +\gamma \right)\sin \left( \beta -\gamma \right)\], we get
\[\begin{align}
  & 2\cos \left( \beta +\gamma \right)\sin \left( \beta -\gamma \right) \\
 & =\sin \left( \beta +\gamma +\beta -\gamma \right)-\sin \left( \beta +\gamma -\beta +\gamma \right) \\
 & =\sin \left( 2\beta \right)-\sin \left( 2\gamma \right) \\
\end{align}\]
For \[2\cos \left( \alpha +\gamma \right)\sin \left( \gamma -\alpha \right)\], we get
\[\begin{align}
  & 2\cos \left( \alpha +\gamma \right)\sin \left( \gamma -\alpha \right) \\
 & =\sin \left( \alpha +\gamma +\gamma -\alpha \right)-\sin \left( \alpha +\gamma -\gamma +\alpha \right) \\
 & =\sin \left( 2\gamma \right)-\sin \left( 2\alpha \right) \\
\end{align}\]
For \[2\cos \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)\], we get
\[\begin{align}
  & 2\cos \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right) \\
 & =\sin \left( \alpha +\beta +\alpha -\beta \right)-\sin \left( \alpha +\beta -\alpha +\beta \right) \\
 & =\sin \left( 2\alpha \right)-\sin \left( 2\beta \right) \\
\end{align}\]
Therefore,
 $ \begin{align}
  & \cos \alpha \sin \left( \beta -\gamma \right)+\cos \beta \sin \left( \gamma -\alpha \right)+\cos \gamma \sin \left( \alpha -\beta \right) \\
 & =-\dfrac{1}{2}\left[ \sin \left( 2\beta \right)-\sin \left( 2\gamma \right)+\sin \left( 2\gamma \right)-\sin \left( 2\alpha \right)+\sin \left( 2\alpha \right)-\sin \left( 2\beta \right) \right] \\
 & =0 \\
\end{align} $
The correct option is A.
So, the correct answer is “Option A”.

Note: The trigonometric functions of multiple angles are the multiple angle formula. Double and triple angles formulas are there under the multiple angle formulas. Sine, tangent and cosine are the general functions for the multiple angle formula. Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $ -\infty \le x\le \infty $ . In that case we have to use the formula $ x=n\pi \pm a $ for $ \cos \left( x \right)=\cos a $ where $ 0\le a\le \pi $ .