JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

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

    A) 0

    B) 1

    C) -1

    D) 2

    Correct Answer: C

    Solution :

    [c] : We have, \[\sin x+{{\sin }^{2}}x=1\] or\[\sin x=1-{{\sin }^{2}}x\]or\[\sin x={{\cos }^{2}}x\] \[\therefore \]\[{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x-2\] \[={{\sin }^{6}}x+3{{\sin }^{5}}x+3{{\sin }^{4}}x+{{\sin }^{3}}x-2\] \[={{(si{{n}^{2}}x)}^{3}}+3{{(si{{n}^{2}}x)}^{2}}\sin x+3(si{{n}^{2}}x){{(sinx)}^{2}}\] \[+{{(sinx)}^{3}}-2\] \[={{(si{{n}^{2}}x+sinx)}^{3}}-2={{(1)}^{3}}-2=-1\].

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