JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of multiple and sub multiple angles

  • question_answer
    \[\frac{{{\cot }^{2}}15{}^\circ -1}{{{\cot }^{2}}15{}^\circ +1}=\] [MP PET 1998]

    A) \[\frac{1}{2}\]

    B) \[\frac{\sqrt{3}}{2}\]

    C) \[\frac{3\sqrt{3}}{4}\]

    D) \[\sqrt{3}\]

    Correct Answer: B

    Solution :

    \[\frac{{{\cot }^{2}}{{15}^{o}}-1}{{{\cot }^{2}}{{15}^{o}}+1}=\frac{\frac{{{\cos }^{2}}{{15}^{o}}}{{{\sin }^{2}}{{15}^{o}}}-1}{\frac{{{\cos }^{2}}{{15}^{o}}}{{{\sin }^{2}}{{15}^{o}}}+1}\] \[=\frac{{{\cos }^{2}}{{15}^{o}}-{{\sin }^{2}}{{15}^{o}}}{{{\cos }^{2}}{{15}^{o}}+{{\sin }^{2}}{{15}^{o}}}=\cos ({{30}^{o}})=\frac{\sqrt{3}}{2}\].


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