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this is a question asked by my iit teacher as a challenge..... so I need ur utmost help to get it`s answer...Please answer me as soon as possible

this is a question asked by my iit teacher as a challenge..... so I need ur utmost help to get it`s answer...Please answer me as soon as possible

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Grade:11

1 Answers

Arun
25750 Points
6 years ago
because sin\theta lies between [ –1,1]
hence sin^2\theta 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
if found any problem, do comment here

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