A) 0
B) 1
C) 2
D) \[\frac{3}{2}\]
E) \[\frac{2}{3}\]
Correct Answer: C
Solution :
\[\because \]\[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1\] \[\therefore \]\[1-{{\sin }^{2}}\alpha +1-{{\sin }^{2}}\beta +1-{{\sin }^{2}}\gamma =1\] \[\Rightarrow \] \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma =2\]You need to login to perform this action.
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