A) 0
B) 1
C) 2
D) \[6/\sqrt{2}\]
Correct Answer: C
Solution :
(c): \[\sin A+\frac{1}{\operatorname{sinA}}=2\] \[\Rightarrow \] \[{{\sin }^{2}}A+1=2\sin A\] \[\Rightarrow \] \[{{\sin }^{2}}A-2\sin A+1=0\] \[\Rightarrow \] \[{{(sinA-1)}^{2}}=0\] \[\Rightarrow \] \[\sin A=1\] \[\therefore \] \[\frac{{{\sin }^{4}}A+1}{{{\sin }^{2}}A}=\frac{1+1}{1}=2\] Again here, you can straightaway conclude that \[\sin A=1\] \[\therefore \] \[\frac{{{\sin }^{4}}A+1}{{{\sin }^{2}}A}=\frac{1+1}{1}=2\]
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