A) \[~x=y\]
B) \[x=-y\]
C) \[2x=y\]
D) none of these
Correct Answer: A
Solution :
\[{{\sin }^{2}}\theta =\frac{{{x}^{2}}+{{y}^{2}}}{2xy}\] \[\Rightarrow \]\[{{(\pm \,1)}^{2}}=\frac{{{x}^{2}}+{{y}^{2}}}{2xy}\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}=2xy\] \[{{x}^{2}}+{{y}^{2}}-2xy=0\Rightarrow {{(x-y)}^{2}}=0\Rightarrow y=x\]You need to login to perform this action.
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