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

  • question_answer
    \[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}\]is equal to

    A) \[\tan (A-B)\]     

    B) \[\tan (A+B)\]

    C) \[cot(A-B)\]       

    D) \[cot(A+B)\]

    Correct Answer: B

    Solution :

    [b] \[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}\] \[=\,\frac{2\sin (A+B)sin(A-B)}{\sin 2A-\sin 2B}\] \[=\,\frac{2\sin (A+B)sin(A-B)}{2\sin (A-B)cos(A+B)}\] \[=\tan (A+B)\]


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