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Question-215679




Question Number 215679 by BaliramKumar last updated on 14/Jan/25
Answered by MATHEMATICSAM last updated on 14/Jan/25
sec^2 θ = ((4xy)/((x + y)^2 ))  ⇒ cos^2 θ = (((x + y)^2 )/(4xy))  (x − y)^2  ≥ 0   ⇒ (x + y)^2  − 4xy ≥ 0  ⇒ (x + y)^2  ≥ 4xy  ⇒ (((x + y)^2 )/(4xy)) ≥ 1  and its equal to 1 when x = y  We know cos^2 θ ≤ 1  So cos^2 θ will be (((x + y)^2 )/(4xy)) when x = y.  So sec^2 θ = ((4xy)/((x + y)^2 )) is only possible when  x = y.
sec2θ=4xy(x+y)2cos2θ=(x+y)24xy(xy)20(x+y)24xy0(x+y)24xy(x+y)24xy1anditsequalto1whenx=yWeknowcos2θ1Socos2θwillbe(x+y)24xywhenx=y.Sosec2θ=4xy(x+y)2isonlypossiblewhenx=y.
Answered by A5T last updated on 14/Jan/25
(x+y)^2 ≥4xy⇒sec^2 θ=((4xy)/((x+y)^2 ))≤((4xy)/(4xy))=1  But sec^2 θ=(1/(cos^2 θ))≥1   ⇒sec^2 θ≥1 and sec^2 θ≤1 which is only possible  when secθ=1⇒4xy=(x+y)^2   ⇒x^2 −2xy+y^2 =(x−y)^2 =0⇒x=y
(x+y)24xysec2θ=4xy(x+y)24xy4xy=1Butsec2θ=1cos2θ1sec2θ1andsec2θ1whichisonlypossiblewhensecθ=14xy=(x+y)2x22xy+y2=(xy)2=0x=y

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