Question

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

• A. 2 sin α sin β cos γ
• B. 2 cos α sin β cos γ
• C. 2 sin α sin β sin γ
• D. Always 0

Answer 1

The correct option is A

2 sin α sin β cos γ

We have, $\alpha +\beta -\gamma =\pi$
$Now,si{n}^2\alpha +si{n}^2\beta -si{n}^2\gamma =si{n}^2\alpha +sin(\beta -\gamma )sin(\beta +\gamma )=si{n}^2\alpha +sin(\pi -\alpha )sin(\beta +\gamma )=si{n}^2\alpha +sin\alpha sin(\beta +\gamma )=sin\alpha [sin\alpha +sin(\beta -\gamma \left)\right]=sin\alpha \left[sin\right\{\pi -(\beta -\gamma )\}+sin(\beta +\gamma \left)\right]=sin\alpha \left[sin\right(\beta -\gamma )+sin(\beta +\gamma \left)\right]=sin\alpha 2sin\beta cos\gamma =2sin\alpha sin\beta cos\gamma$