sec2θ=4xy(x+y)2 is true if and only if
x+y≠0
x=y,x≠0
x =y
x≠0,y≠0
We have:
sec2θ=4xy(x+y)2
⇒4xy(x+y)2≥1[∵sec2θ≥1]
⇒4xy≥(x+y)2
⇒4xy≥x2+y2+2xy
⇒2xy≥x2+y2
⇒(x−y)2≤0
⇒(x−y)≤0
⇒x=y
For x =0, sec2θ will not be defined,
⇒x≠0
∴x=y
if xy > 0, then