Question

\[\sec^2 x = \frac{4xy}{(x + y )^2}\] is true if and only if

 

विकल्प

• x + y ≠ 0

• x = y, x ≠ 0

• x = y

• x ≠0, y ≠ 0

Answer 1

 x = y, x ≠ 0

We have:

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

\[ \Rightarrow \frac{4xy}{(x + y )^2} \geq 1 \left[ \because \sec^2 x \geq 1 \right]\]

\[ \Rightarrow 4xy\geq(x + y )^2\]
\[\Rightarrow 4xy \geq x^2 + y^2 + 2xy\]
\[ \Rightarrow 2xy \geq x^2 + y^2 \]
\[ \Rightarrow \left( x - y \right)^2 \leq 0\]
\[ \Rightarrow \left( x - y \right) \leq 0\]
\[ \Rightarrow x = y\]
\[\text{ For }x = 0, \sec^2 x \text{ will not be defined,} \]
\[ \Rightarrow x \neq 0\]
\[ \therefore x = y\]