because sin
lies between [ –1,1]
hence sin^2
will lie between [0,1]
now 0
1st part-
(x^2+y^2)/2xy >= 0
(x^2+y^2) >= 0
2nd part-
(x^2+y^2)/2xy
(x^2+y^2)
(x^2+y^2- 2xy)
(x –y)^2
now square of anything can’t be less than 0
hence (x –y)^2 = 0
x-y = 0
x= y this will be condition (this is your answer)
hence x=y = any value except 0
because at x,y= 0, function will be not defined
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