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Question

sec2θ=4xy(x+y)2 is true if and only if
(a) x + y ≠ 0
(b) x = y, x ≠ 0
(c) x = y
(d) x ≠0, y ≠ 0

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Solution

(b) x = y, x ≠ 0

We have:sec2θ=4xy(x+y)24xy(x+y)21 sec2θ14xy (x+y)2
4xy x2+y2+2xy2xy x2+y2x-y2 0x-y 0 x= yFor x = 0, sec2θ will not be defined, x 0 x=y

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