If sin theta=(x+y)^2/4xy then prove that x=y
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wrong question
it is incorrect
it might be like this
max value of
sintheta = 1
(x+y)^2/4xy = 1
(x+y)^2 = 4xy
(x-y)^2 = 0
x-y =0
x = y
it is incorrect
it might be like this
max value of
sintheta = 1
(x+y)^2/4xy = 1
(x+y)^2 = 4xy
(x-y)^2 = 0
x-y =0
x = y
bidhanchsahanap50ub7:
then solve it correcting.....can it be sin^2 thetha
see it
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