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Q.
The equation $\sin ^{2} \theta=\frac{x^{2}+y^{2}}{2 x y}$ is possible if
Trigonometric Functions
Solution:
We have $\frac{x^{2}+y^{2}}{2 x y}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$
Now, $\sin ^{2} \theta=\frac{x^{2}+y^{2}}{2 x y}$
$\Rightarrow \frac{x^{2}+y^{2}}{2 x y} \geq 0$
$\left[\because \sin ^{2} \theta \geq 0\right]$
Therefore, $x$ and $y$ have the same sign.
Now, $\frac{x^{2}+y^{2}}{2 x y}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$
$\Rightarrow \frac{x^{2}+y^{2}}{2 x y} \geq 1$
$(\because A.M. \geq G.M.)$
But $\sin ^{2} \theta \leq 1$. Therefore, $\frac{x^{2}+y^{2}}{2 x y}=1$
$\Rightarrow x=y$.