A) 0
B) 1
C) - 1
D) 2
Correct Answer: C
Solution :
We have, \[\sin x+{{\sin }^{2}}x=1\] or \[\sin x=1-{{\sin }^{2}}x\] or \[\sin x={{\cos }^{2}}x\] \ \[{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x-2\] \[={{\sin }^{6}}x+3{{\sin }^{5}}x+3{{\sin }^{4}}x+{{\sin }^{3}}x-2\] \[={{({{\sin }^{2}}x)}^{3}}+3{{({{\sin }^{2}}x)}^{2}}\sin x\] \[+3({{\sin }^{2}}x){{(\sin x)}^{2}}+{{(\sin x)}^{3}}-2\] \[={{({{\sin }^{2}}x+\sin x)}^{3}}-2\]\[={{(1)}^{3}}-2\] \[[\because \sin x+{{\sin }^{2}}x=1(\text{given})]\] = - 1.
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