CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    \[{{\cos }^{-1}}\left( \frac{3+5\cos x}{5+3\cos x} \right)\]is equal to:

    A)  \[{{\tan }^{-1}}\left( \frac{1}{2}\tan \frac{x}{2} \right)\]

    B)  \[2{{\tan }^{-1}}\left( 2\tan \frac{x}{2} \right)\]

    C)  \[\frac{1}{2}{{\tan }^{-1}}\left( 2\tan \frac{x}{2} \right)\]

    D)  \[2{{\tan }^{-1}}\left( \frac{1}{2}\tan \frac{x}{2} \right)\]

    E)  \[{{\tan }^{-1}}\left( \tan \frac{x}{2} \right)\]

    Correct Answer: D

    Solution :

    \[{{\cos }^{-1}}\left( \frac{3+5\cos x}{5+3\cos x} \right)\] \[={{\cos }^{-1}}\left\{ \frac{3+5\left( \frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2} \right)}{5+3\left( \frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2} \right)} \right\}\] \[={{\cos }^{-1}}\left( \frac{4-{{\tan }^{2}}x/2}{4+{{\tan }^{2}}x/2} \right)\] \[={{\cos }^{-1}}\left\{ \frac{1-{{\left( \frac{1}{2}\tan x/2 \right)}^{2}}}{1+{{\left( \frac{1}{2}\tan x/2 \right)}^{2}}} \right\}\] \[=2{{\tan }^{-1}}\left( \frac{1}{2}\tan \frac{x}{2} \right)\]           


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