CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    \[{{\sin }^{2}}{{17.5}^{o}}+{{\sin }^{2}}{{72.5}^{o}}\] is equal to

    A)  \[{{\cos }^{2}}{{90}^{o}}\]                          

    B)  \[{{\tan }^{2}}{{45}^{o}}\]

    C)  \[co{{s}^{2}}{{30}^{o}}\]                             

    D)  \[{{\sin }^{2}}{{45}^{o}}\]

    Correct Answer: B

    Solution :

    We have, \[{{\sin }^{2}}{{17.5}^{o}}+{{\sin }^{2}}{{72.5}^{o}}\] \[={{\sin }^{2}}{{17.5}^{o}}+{{\sin }^{2}}({{90}^{o}}-{{17.5}^{o}})\] \[={{\sin }^{2}}{{17.5}^{o}}+{{\cos }^{2}}{{17.5}^{o}}\] \[=1={{1}^{2}}\] \[={{\tan }^{2}}{{45}^{o}}\]


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