JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Solution of trigonometrical equations

  • question_answer
    If \[\sin (A+B)\]=1 and \[\cos (A-B)=\frac{\sqrt{3}}{2},\]then the smallest positive values of A and B are

    A) \[{{60}^{o}},\text{ }{{30}^{o}}\]

    B) \[{{75}^{o}},\text{ }{{15}^{o}}\]

    C) \[{{45}^{o}},\text{ }{{60}^{o}}\]

    D) \[{{45}^{o}},\text{ }{{45}^{o}}\]

    Correct Answer: A

    Solution :

    \[\sin (A+B)=1\] and\[\cos (A-B)=\frac{\sqrt{3}}{2}\] \[\Rightarrow \] \[A+B=\frac{\pi }{2}\] and \[A-B=\frac{\pi }{6}\] \[\Rightarrow \] \[A=\frac{\pi }{3},\,B=\frac{\pi }{6}\].


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