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. \[\therefore \,\,A=B=C={{60}^{0}}\] \[\Rightarrow \,\,{{\sin }^{2}}A+{{\cos }^{2}}B+{{\tan }^{2}}C\]\[=\frac{3}{4}+\frac{1}{4}+3=4\]You need to login to perform this action.
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