JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2004

  • question_answer
        In a \[\frac{1}{27}\left( \begin{matrix}    1 & -26  \\    0 & -27  \\ \end{matrix} \right)\], if\[\frac{1}{27}\left( \begin{matrix}    -1 & -26  \\    0 & -27  \\ \end{matrix} \right)\], then the value of\[\alpha +\beta -\lambda =\pi \]is equal to :

    A)  1                                            

    B)  2

    C)  \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta -{{\sin }^{2}}\gamma \]                               

    D)  \[\text{2 sin }\alpha \text{ sin}\beta \text{ cos }\gamma \]

    Correct Answer: B

    Solution :

                    \[\cot \frac{B}{2}\cot \frac{C}{2}\] \[=\sqrt{\frac{s(s-b)}{(s-a)(s-c)}.\frac{s(s-c)}{(s-a)(s-b)}}\] \[=\frac{s}{(s-a)}\] [\[\because \]\[3a=b+c\text{ }or\text{ a}+b+c=2s=4a\]] \[=\frac{2a}{a}=2\]


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