A) $ \frac{1}{4} $
B) $ \frac{1}{4}(1+6x^{2}) $
C) $ \frac{1}{4}(1+6x^{2}-3x^{4}) $
D) $ \frac{1}{2}(5-3x^{2}) $
Correct Answer: C
$ \therefore $ $ \sin \theta \cos \theta =\frac{x^{2}-1}{2} $ $ {{\cos }^{6}}\theta +{{\sin }^{6}}\theta ={{({{\cos }^{2}}\theta )}^{3}}+{{({{\sin }^{2}}\theta )}^{3}} $ $ =({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )(cos^{4}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{4}}\theta ) $ $ =[{{(cos^{2}\theta )}^{2}}+{{({{\sin }^{2}}\theta )}^{2}}-{{\cos }^{2}}\theta {{\sin }^{2}}\theta ] $ $ =[{{(cos^{2}\theta +{{\sin }^{2}}\theta )}^{2}}-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta ] $ $ =1-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta =1-3{{( \frac{x^{2}-1}{2} )}^{2}} $ $ =1-\frac{3(x^{4}-2x^{2}+1)}{4}=\frac{4-3x^{4}+6x^{2}-3}{4} $ $ =\frac{1-3x^{4}+6x^{2}}{4}=\frac{1}{4}(1+6x^{2}-3x^{4}) $