Question

The equation ${\sin }^2\theta =\frac{x^2+y^2}{2xy}$ is possible if

• A. $x=y$
• B. $x=-y$
• C. $2x=y$
• D. none of these

Answer 1

Answer: (A)

We have $\frac{x^2+y^2}{2xy}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$
Now, ${\sin }^2\theta =\frac{x^2+y^2}{2xy}$
$\Rightarrow \frac{x^2+y^2}{2xy}\ge 0$
$\left[\because {\sin }^2\theta \ge 0\right]$
Therefore, $x$ and $y$ have the same sign.
Now, $\frac{x^2+y^2}{2xy}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$
$\Rightarrow \frac{x^2+y^2}{2xy}\ge 1$
$(\because A.M.\ge G.M.)$
But ${\sin }^2\theta \le 1$. Therefore, $\frac{x^2+y^2}{2xy}=1$
$\Rightarrow x=y$.