A) \[0\]
B) \[1\]
C) \[-1\]
D) \[2\]
Correct Answer: B
Solution :
Given that \[\sin x+{{\sin }^{2}}x=1\] \[\Rightarrow \] \[\sin x=1-{{\sin }^{2}}x\] \[\sin x={{\cos }^{2}}x\] ... (i) \[\therefore \]\[{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x\] \[={{({{\cos }^{2}}x)}^{6}}+3{{({{\cos }^{2}}x)}^{5}}+3{{({{\cos }^{2}}x)}^{4}}\] \[+{{({{\cos }^{2}}x)}^{3}}\] \[={{\sin }^{6}}x+3{{\sin }^{5}}x+3{{\sin }^{4}}x+{{\sin }^{3}}x\] [from (i)] \[={{({{\sin }^{2}}x+\sin x)}^{3}}\] \[=1\]
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