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JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    In triangle ABC, if \[\left| \begin{matrix}  1 & a & b  \\    1 & c & b  \\    1 & b & c  \\ \end{matrix} \right|=0\] then the value of \[{{\sin }^{2}}A+{{\cos }^{2}}B+{{\tan }^{2}}C\] is equal to

    A)  2                                

    B)  4

    C)  \[\frac{9}{2}\]                                     

    D)  \[\frac{11}{2}\]

    Correct Answer: B

    Solution :

    \[\left| \begin{matrix}    0 & a-c & b-c  \\    0 & c-b & a-c  \\    1 & b & c  \\ \end{matrix} \right|\ =0\] \[{{(a-c)}^{2}}+(b-c)(b-a)=0\] \[{{a}^{2}}+{{c}^{2}}-ac\,+{{b}^{2}}-ab-bc=0\] \[\therefore \,\,a=b=c\Rightarrow \] triangle is equilateral. \[\therefore \,\,A=B=C={{60}^{0}}\] \[\Rightarrow \,\,{{\sin }^{2}}A+{{\cos }^{2}}B+{{\tan }^{2}}C\]\[=\frac{3}{4}+\frac{1}{4}+3=4\]


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