SSC Quantitative Aptitude Trigonometry Question Bank Trigonometry (II)

  • question_answer
    If \[\tan \theta -\cot \theta =0\]and \[\theta \] is positive acute angle, then the value of \[\frac{\tan \,(\theta +15{}^\circ )}{\tan \,(\theta -15{}^\circ )}\] is [SSC CGL Tier II, 2015]

    A) \[\sqrt{3}\]

    B) \[\frac{1}{\sqrt{3}}\]

    C) \[3\]

    D) \[\frac{1}{3}\]

    Correct Answer: C

    Solution :

    [c] \[\tan \theta -\cot \theta =0\] \[\Rightarrow \]\[\tan \theta =\cot \theta \]\[\Rightarrow \]\[\tan \theta =\tan (90{}^\circ -\theta )\]\[\Rightarrow \]\[\theta =90{}^\circ -\theta \] \[\Rightarrow \]   \[2\theta =90{}^\circ \] \[\therefore \]      \[\theta =45{}^\circ \] Then, \[\frac{\tan (\theta +15{}^\circ )}{\tan (\theta -15{}^\circ )}=\frac{\tan (45{}^\circ +15{}^\circ )}{\tan (45{}^\circ -15{}^\circ )}=\frac{\tan 60{}^\circ }{\tan 30{}^\circ }=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=3\]


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