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10th Class Mathematics Introduction to Trigonometry Question Bank MCQs - Introduction to Trigonometry

  • question_answer
    If \[\frac{{{\tan }^{3}}\theta -1}{\tan \theta -1}=A\,{{\sec }^{2}}\theta +B\tan \theta ,\] then \[A+B\] is equal to:

    A) \[1\]

    B) \[-1\]

    C) \[2\]

    D) \[3\]

    Correct Answer: C

    Solution :

    [c]\[\frac{{{\tan }^{3}}\theta -1}{\tan \theta -1}=\frac{(\tan \theta -1)({{\tan }^{2}}\theta +1+\tan \theta )}{(\tan \theta -1)}\]
    \[={{\sec }^{2}}\theta +\tan \theta \]\[[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ]\]
    \[=A{{\sec }^{2}}\theta +B\tan \theta \]
    On comparing, we get
    \[A=1\] and \[B=1\]
    \[\therefore \,\,\,\,\,\,\,\,\,\,A+B=1+1=2\]


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