A) \[a=1,b=6\]
B) \[a=1,b=-6\]
C) \[a=6,b=1\]
D) None of these
Correct Answer: B
Solution :
\[\tan 45{}^\circ =\frac{2\sqrt{\frac{1}{4}-ab}}{a+b}\] \[\Rightarrow {{(a+b)}^{2}}=(1-4ab)\Rightarrow {{a}^{2}}+{{b}^{2}}+6ab-1=0\], which is obviously satisfied by \[a=1\]and \[b=-6\].
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