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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    If A ,B and C have positive values such that \[A+B+C=\pi ;\] then the minimum value of \[\frac{\cos \left( \frac{A-B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)}+\frac{\cos \left( \frac{B-C}{2} \right)}{\cos \left( \frac{B+C}{2} \right)}+\frac{\cos \left( \frac{C-A}{2} \right)}{\cos \left( \frac{C+A}{2} \right)}A\]is

    A) \[3\]               

    B)        \[6\]            

    C) \[9\]                      

    D)        \[12\]

    Correct Answer: B

    Solution :

    [b] \[\frac{\cos \left( \frac{A-B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)}=\frac{2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right)}{2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A+B}{2} \right)}\] \[=\frac{\sin A}{\sin C}+\frac{\sin B}{\sin C}\] \[\therefore \,\,\frac{\cos \left( \frac{A-B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)}+\frac{\cos \left( \frac{B-C}{2} \right)}{\cos \left( \frac{B+C}{2} \right)}+\frac{\cos \left( \frac{C-A}{2} \right)}{\cos \left( \frac{C+A}{2} \right)}\] \[=\left( \frac{\sin A}{\sin C}+\frac{\sin B}{\sin C} \right)+\left( \frac{\sin B}{\sin A}+\frac{\sin C}{\sin A} \right)+\left( \frac{\sin C}{\sin B}+\frac{\sin A}{\sin B} \right)\]\[=\left( \frac{\sin A}{\sin C}+\frac{\sin C}{\sin A} \right)+\left( \frac{\sin B}{\sin C}+\frac{\sin C}{\sin B} \right)+\left( \frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \right)\ge 6\]\[\left\{ \because \,\,for\,t>0,\,t+\frac{1}{t}\ge 2 \right\}\]         


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