JEE Main & Advanced Mathematics Inverse Trigonometric Functions Question Bank Mock Test - Inverse Trigonometric Functions

  • question_answer
    If \[{{\tan }^{-1}}(si{{n}^{2}}\theta -2sin\theta +3)+co{{t}^{-1}}({{5}^{{{\sec }^{2}}y}}+1)=\frac{\pi }{2}\], then the value of \[{{\cos }^{2}}\theta -\sin \theta \]is equal to

    A) 0                     

    B) -1

    C) 1                     

    D) none of these

    Correct Answer: C

    Solution :

    [c] From the given equation \[{{\sin }^{2}}\theta -2\sin \theta +3={{5}^{{{\sec }^{2}}y}}+1\], we get \[{{(sin\theta -1)}^{2}}+2={{5}^{{{\sec }^{2}}y}}+1\] \[\text{L}\text{.}H.S.\le 6,R.H.S.\ge 6\] Possible solution is \[\sin \theta =-1\] when L.H.S. = R.H.S. \[\Rightarrow {{\cos }^{2}}\theta =0\] \[\Rightarrow {{\cos }^{2}}\theta -\sin \theta =1\]


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