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Question

sec2θ=4xy(x+y)2 is true if and only if
  1. x+y0
  2. x=y,x0
  3. x=y
  4. x0,y0

A
x=y,x0
B
x+y0
C
x=y
D
x0,y0
Solution
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sec2θ1

4xy(x+y)21(x+y)24xy0

(xy)20(xy)2=0

Since (xy)2 can never be negative

x=y

Here,the fraction 4xy(x+y)2 is not defined for

x+y=0

x+y0x02x0 or x0

Hence x=y,x0

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