JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        In a triangle ABC, \[2ca\sin \frac{A-B+C}{2}\] is equal to:

    A)  \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}\]              

    B) \[{{c}^{2}}+{{a}^{2}}-{{b}^{2}}\]

    C)  \[{{b}^{2}}-{{c}^{2}}-{{a}^{2}}\]               

    D)  \[{{c}^{2}}-{{a}^{2}}-{{b}^{2}}.\]

    Correct Answer: B

    Solution :

                     We have\[A+C=\pi -B\] and \[\frac{A-B+C}{2}=\frac{\pi }{2}-B\] \[\therefore \]  \[2ac\cos B=2ca\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ca}\]                 \[={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]


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