Question

${\sec }^{2}x=\frac{4xy}{(x+y{)}^2}$ is true if and only if
(a) x + y ≠ 0
(b) x = y, x ≠ 0
(c) x = y
(d) x ≠0, y ≠ 0

Answer 1

(b) x = y, x ≠ 0

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ {\sec }^2\mathrm{x}=\frac{4xy}{(x+y{)}^2} \\ \Rightarrow \frac{4xy}{(x+y{)}^2}\ge 1\left[\because {\sec }^2\mathrm{x}\ge 1\right] \\ \Rightarrow 4xy\ge (x+y{)}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow 4xy\ge x^2+y^2+2xy \\ \Rightarrow 2xy\ge x^2+y^2 \\ \Rightarrow {\left(x-y\right)}^2\le 0 \\ \Rightarrow \left(x-y\right)\le 0 \\ \Rightarrow x=y \\ \mathrm{For}x=0,se{c}^{2}x\mathrm{will}\mathrm{not}\mathrm{be}\mathrm{defined}, \\ \Rightarrow x\ne 0 \\ \therefore x=y \end{gathered}$$