A) \[\frac{1}{6}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{12}\]
Correct Answer: D
Solution :
\[\frac{1}{4}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)-\frac{1}{6}\left( {{\sin }^{6}}x+{{\cos }^{6}}x \right)\] \[=\frac{3({{\sin }^{4}}x+{{\cos }^{4}}x)-2\left( {{\sin }^{6}}x+{{\cos }^{6}}x \right)}{12}\] \[=\frac{3\left( 1-2{{\sin }^{2}}x{{\cos }^{2}}x \right)-2\left( 1-3{{\sin }^{2}}x{{\cos }^{2}}x \right)}{12}\] \[=\frac{1}{12}.\]
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