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SSC Quantitative Aptitude Trigonometry Question Bank Trigonometry (II)

  • question_answer
    If \[\tan (A+B)=\sqrt{3}\]and \[\tan (A+B)=\frac{1}{\sqrt{3}},\] where A B and (A + B) is acute, then A is equal to

    A) \[15{}^\circ \]

    B) \[30{}^\circ \]

    C) \[45{}^\circ \]

    D) \[60{}^\circ \]

    Correct Answer: C

    Solution :

    [c] \[\tan \,(A+B)=\sqrt{3}=\tan 60{}^\circ \] \[\Rightarrow \]   \[A+B=60{}^\circ \] and \[\tan \,(A-B)=\frac{1}{\sqrt{3}}=\tan \text{ }30{}^\circ \] \[\Rightarrow \]   \[A-B=30{}^\circ \]…(ii) On adding Eqs. (i) and (ii), we get \[2A=90{}^\circ \] \[\Rightarrow \]   \[A=45{}^\circ \]


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