A) \[\frac{3}{5}\]
B) \[\frac{2}{3}\]
C) \[\frac{1}{5}\]
D) None of these
Correct Answer: A
Solution :
Here, \[l=\cos \theta ,\,m=\cos \beta ,\,n=\cos \theta \], \[(\because \,\,l=n)\] Now, \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\Rightarrow 2{{\cos }^{2}}\theta +{{\cos }^{2}}\beta =1\] \[\Rightarrow \] \[2{{\cos }^{2}}\theta ={{\sin }^{2}}\beta \] Given, \[{{\sin }^{2}}\beta =3{{\sin }^{2}}\theta \] Þ \[2{{\cos }^{2}}\theta =3{{\sin }^{2}}\theta \] Þ \[5{{\cos }^{2}}\theta =3,\,\therefore \,{{\cos }^{2}}\theta =\frac{3}{5}.\]
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