The equation $\cos ^{2} \theta=\frac{(x+y)^{2}}{4 x y}$ is only possible when ?
A. x = -y B. x > y C. x = y D. x < y Answer: Option CShow Answer
Solution(By Apex Team)
$\begin{array}{l}\cos^2\theta=\frac{(x+y)^2}{4xy}\\
\text{Max value of cos}^2\theta=1\\
\begin{array}{l}
\Rightarrow 1=\frac{(x+y)^{2}}{4 x y} \\
\Rightarrow 4 x y=(x+y)^{2} \\
\Rightarrow 4 x y=x^{2}+y^{2}+2 x y \\
\Rightarrow 0=x^{2}+y^{2}-2 x y \\
\Rightarrow 0=(x-y)^{2} \\
\Rightarrow 0=x-y \\
\Rightarrow x=y
\end{array}\end{array}$
