The equation sin^2theta=(x^2+y^2)/(2x y)...

The equation `sin^2theta=(x^2+y^2)/(2x y),x , y!=0` is possible if

x = y

x = - y

2x = y

None of these

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The correct Answer is:

Now, `sin^2theta=(x^2+y^2)/(2xy)`
`:. ` x,y have same sign
Now `(x^2+y^2)/(2xy)=1/2[(sqrt(x/y)-sqrt(y/x))^2+2]ge1`
But `sinthetale1," Therefore", (x^2+y^2)/(2xy)=1`
`rArrx=y`.

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