sec² x = 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

Question

\[ \sec^2x=\frac{4xy}{(x+y)^2} \]

\[ \text{is true if and only if} \]

(a) \(x+y\neq0\)
(b) \(x=y,\ x\neq0\)
(c) \(x=y\)
(d) \(x\neq0,\ y\neq0\)

Since

\[ \sec^2x\ge1 \]

So,

\[ \frac{4xy}{(x+y)^2}\ge1 \]

\[ 4xy\ge(x+y)^2 \]

\[ 4xy\ge x^2+2xy+y^2 \]

\[ 0\ge x^2-2xy+y^2 \]

\[ 0\ge(x-y)^2 \]

But

\[ (x-y)^2\ge0 \]

Hence,

\[ (x-y)^2=0 \]

\[ x=y \]

Also, denominator should not be zero:

\[ x+y\neq0 \]

Since \(x=y\),

\[ 2x\neq0 \Rightarrow x\neq0 \]

Therefore,

\[ x=y,\ x\neq0 \]

\[ \boxed{x=y,\ x\neq0} \]

Correct Option: (b)

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