Question

If $\alpha +\beta +\gamma =\pi$, then ${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma$ is equal to

• A. $2\sin \alpha \sin \beta \cos \gamma$
• B. $2\cos \alpha \cos \beta \cos \gamma$
• C. $2\sin \alpha \sin \beta \sin \gamma$
• D. None of these

Answer 1

The correct option is A $2\sin \alpha \sin \beta \cos \gamma$

${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma ={\sin }^2\alpha +\sin (\beta +\gamma )\sin (\beta -\gamma )={\sin }^2\alpha +\sin (\pi -\alpha )\sin (\beta -\gamma )={\sin }^2\alpha +\sin \alpha \sin (\beta -\gamma )=\sin \alpha (\sin \alpha +\sin (\beta -\gamma ))=\sin \alpha (\sin (\pi -(\beta +\gamma ))+\sin (\beta -\gamma ))=\sin \alpha (\sin (\beta +\gamma )+\sin (\beta -\gamma ))=\sin \alpha \left(2\sin \beta \cos \gamma \right)=2\sin \alpha \sin \beta \cos \gamma$