A) \[xy\]
B) \[-xy\]
C) \[A=\left[ \begin{matrix} 3 & 2 \\ 0 & 1 \\ \end{matrix} \right]\]
D) none of these
Correct Answer: A
Solution :
We have, \[\alpha +\beta -\gamma =\pi \] Now, \[si{{n}^{2}}\alpha +si{{n}^{2}}\beta -si{{n}^{2}}\gamma \] \[=si{{n}^{2}}\alpha +sin(\beta -\gamma )+sin(\beta +\gamma )\] \[=si{{n}^{2}}\alpha +sin(\pi -\alpha )+sin(\beta +\gamma )\] \[[\because \alpha +\beta -\gamma =\pi ]\] \[={{\sin }^{2}}\alpha +\sin \alpha \sin (\beta +\gamma )\] \[=\sin \alpha [\sin \alpha +\sin (\beta +\gamma )]\] \[=\sin \alpha \{\sin (\pi -\beta +\gamma )+\sin (\beta +\gamma )\}\] \[=\sin \alpha \{-\sin (\gamma -\beta )+\sin (\gamma +\beta )\}\] \[=\sin \alpha \{2\sin \beta \cos \gamma \}\] \[=2\sin \alpha \sin \beta \cos \gamma \]
You need to login to perform this action.
You will be redirected in
3 sec
