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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    If \[\sin x+{{\sin }^{2}}x=1\], then\[{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x\] is equal to:

    A)  1                                            

    B)  2

    C)  3                                            

    D)  0

    Correct Answer: A

    Solution :

    Given                   \[\sin x+{{\sin }^{2}}x=1\] \[\Rightarrow \]               \[\sin x=1-{{\sin }^{2}}x\] \[\Rightarrow \]               \[\sin x={{\cos }^{2}}x\] Now, \[{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x\]                 \[={{\sin }^{6}}x+3{{\sin }^{5}}x+3{{\sin }^{4}}x+{{\sin }^{3}}x\]                 \[=\,{{({{\sin }^{2}}x+\sin x)}^{3}}\]                 \[={{(1)}^{3}}=1\]


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