A) \[2a=b\]
B) \[a=b\]
C) \[a=2b\]
D) None of these
Correct Answer: B
Solution :
We have \[{{(a+b)}^{2}}=4ab{{\sin }^{2}}\theta \] \[\Rightarrow {{\sin }^{2}}\theta =\frac{{{(a+b)}^{2}}}{4ab}\le 1\Rightarrow {{(a+b)}^{2}}-4ab\le 0\] \[\Rightarrow {{(a-b)}^{2}}\le 0\Rightarrow a=b.\]
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