JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer The value of \[co{{s}^{2}}10{}^\circ cos10{}^\circ cos50{}^\circ +co{{s}^{2}}50{}^\circ \]is [JEE Main 9-4-2019 Morning]

    A) \[\frac{3}{2}(1+cos20{}^\circ )\]   

    B) \[\frac{3}{4}\]

    C) \[\frac{3}{4}+\cos {{20}^{o}}\]               

    D) \[\frac{3}{2}\]

    Correct Answer: B

    Solution :

    \[\frac{1}{2}\left( 2{{\cos }^{2}}{{10}^{o}}-2\cos {{10}^{o}}\cos {{50}^{o}}+2{{\cos }^{2}}{{50}^{o}} \right)\] \[\Rightarrow \frac{1}{2}\left( 1+\cos {{20}^{o}}-\left( \cos {{60}^{o}}+\cos {{40}^{o}} \right)+1+\cos {{100}^{o}} \right)\] \[\Rightarrow \frac{1}{2}\left( \frac{3}{2}+\cos {{20}^{o}}+2\sin {{70}^{o}}\sin \left( -{{30}^{o}} \right) \right)\] \[\Rightarrow \frac{1}{2}\left( \frac{3}{2}+\cos {{20}^{o}}-\sin {{70}^{o}} \right)\]\[\Rightarrow \frac{3}{4}\] Ans. [b]


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