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

$2\sin \alpha \sin \beta \cos \gamma$

Find the value of ${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma$:

Given, $\alpha +\beta +\gamma =\pi$

Now,

$\begin{array}{rcl}\sin 2\alpha +\sin 2\beta –\sin 2\gamma & =& \frac{(1–\cos 2\alpha )}{2}+\frac{(1–\cos 2\beta )}{2}–{\sin }^2\gamma \\ & =& \frac{1}{2}+\frac{1}{2}–\frac{1}{2}(\cos 2\alpha +\cos 2\beta )–(1–{\cos }^2\gamma )\\ & =& 1–\frac{1}{2}\left[2\cos \frac{(2\alpha +2\beta )}{2}\cos \frac{(2\alpha –2\beta )}{2}\right]–1+{\cos }^2\gamma \\ & =& –\cos (\alpha +\beta )\cos (\alpha –\beta )+{\cos }^2\gamma \\ & =& -\cos (\pi –\gamma )\cos (\alpha –\beta )+{\cos }^2\gamma \\ & =& \cos \gamma \cos (\alpha –\beta )+{\cos }^2\gamma \\ & =& \cos \gamma \left[\cos \right(\alpha –\beta )+\cos \gamma ]\\ & =& \cos \gamma \left[\cos \right(\alpha –\beta )–\cos (\alpha –\beta \left)\right]\\ & =& \cos \gamma (2\sin \alpha \sin \beta )\\ & =& \mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\alpha }\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\beta }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\gamma }\end{array}$

Hence, Option ‘A’ is Correct.