Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    If\[\cos \theta =-\frac{\sqrt{3}}{2}\]and\[\sin \alpha =-\frac{3}{5}\], where\[\theta \]does not lie in the third quadrant, then \[\frac{2\tan \alpha +\sqrt{3}\tan \theta }{{{\cot }^{2}}\theta +\cos \alpha }\]is equal to

    A) \[\frac{7}{22}\]

    B)                        \[\frac{5}{22}\]

    C) \[\frac{9}{22}\]                

    D)        \[\frac{22}{5}\]

    Correct Answer: B

    Solution :

    Given,\[\cos \theta =-\frac{\sqrt{3}}{2}<0\]and\[\theta \]does not lie in third quadrant. \[\therefore \,\,\theta \]must be lying in 2nd quadrant. \[\Rightarrow \]               \[\tan \theta =-\frac{1}{\sqrt{3}}\] and        \[\cot \theta =-\sqrt{3}\]                                             ... (i) Also,\[\alpha \]lies in 3rd quadrant and\[\sin \alpha =-\frac{3}{5}\] \[\therefore \]  \[\tan \alpha =\frac{3}{4}\]and\[\cos \alpha =-\frac{4}{5}\]         ... (ii) \[\therefore \]  \[\frac{2\tan \alpha +\sqrt{3}\tan \theta }{{{\cot }^{2}}\theta +\cos \alpha }=\frac{2\cdot \frac{3}{4}-\sqrt{3}\cdot \frac{1}{\sqrt{3}}}{3-\frac{4}{5}}\]                 \[=\frac{\frac{3}{2}-1}{3-\frac{4}{5}}=\frac{5}{22}\]


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