CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    \[\frac{\cos 9{}^\circ +\sin 9{}^\circ }{\cos 9{}^\circ -\sin 9{}^\circ }\] is equal to:

    A)  \[tan\text{ }26{}^\circ \]

    B)                         \[tan\text{ 81}{}^\circ \]

    C)  \[tan\text{ 51}{}^\circ \]                             

    D)  \[tan\text{ 54}{}^\circ \]

    E)  \[tan\text{ 46}{}^\circ \]

    Correct Answer: D

    Solution :

    \[\frac{cos\text{ }9{}^\circ +sin\text{ }9{}^\circ }{cos\text{ }9{}^\circ -sin\text{ }9{}^\circ }=\frac{1+tan\text{ }9{}^\circ }{1-tan\text{ }9{}^\circ }\] \[=tan(45{}^\circ +9{}^\circ )\] \[=tan\text{ }54{}^\circ \]


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