Question: The equation $\sec^{2}\theta = \frac{4xy}{(x + y)^{2}}$ is only possible when...

Question

The equation $\sec^{2}\theta = \frac{4xy}{(x + y)^{2}}$ is only possible when

Answer 1

$\because\cos^{2}\theta \leq 1$

⇒ $\sec^{2}\theta = \frac{4xy}{(x + y)^{2}} \geq 1 \Rightarrow 4xy \geq (x + y)^{2} \Rightarrow (x - y)^{2} \leq 0$ Which

is possible only when $x = y$ $(\because x,y \in R)$

Question: The equation \(\sec^{2}\theta = \frac{4xy}{(x + y)^{2}}\) is only possible when...