JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2004

  • question_answer
        The area of triangle formed by the points \[\cot \frac{B}{2}\cot \frac{C}{2}\]is equal to :

    A)  \[\sqrt{3}\]                                       

    B) \[\sqrt{2}\]

    C) \[b\sin A=a,A<\frac{\pi }{2}\]                    

    D)  0

    Correct Answer: D

    Solution :

                    Area\[=\frac{1}{2}\left| \begin{matrix}    a & b+c & 1  \\    b & c+a & 1  \\    c & c+b & 1  \\ \end{matrix} \right|\] \[=\frac{1}{2}\left| \begin{matrix}    a & a+b+c & 1  \\    b & a+b+c & 1  \\    c & a+b+c & 1  \\ \end{matrix} \right|\] applying               \[{{C}_{2}}\to {{C}_{1}}+{{C}_{2}}\]                 \[=\frac{(a+b+c)}{2}\left| \begin{matrix}    a & 1 & 1  \\    b & 1 & 1  \\    c & 1 & 1  \\ \end{matrix} \right|=0\]


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