A) Real
B) Imaginary
C) Greater than 1
D) Equal
Correct Answer: A
Solution :
Given \[{{\sin }^{2}}B=\sin A\cos A\] Þ \[\cos 2B=1-\sin 2A\ge 0\] Now for \[{{x}^{2}}+2x\cot B+1=0\] Consider \[D=4({{\cot }^{2}}B-1)=4\cos 2B\cos \text{e}{{\text{c}}^{\text{2}}}B\ge 0\] Hence, roots are always real.You need to login to perform this action.
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