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a
x+y≠0
b
x=y,x≠0
c
x=y
d
x≠0,y≠0
answer is B.
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Detailed Solution
Given, sec2θ=4xy(x+y)2Now sec2θ≥1⇒4xy(x+y)2≥1or (x+y)2≤4xyor (x+y)2−4xy≤0o r(x−y)2≤0But for real values of x and y, (x−y)2≥0 or (x−y)2=0∴ x=yAlso x+y≠0⇒x≠0.y≠0