If $cot\theta =\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}},$ then what is equal to?
$\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}$
$\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}$
$\dfrac{2xy}{{{x}^{2}}+{{y}^{2}}}$
$\dfrac{2xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$
Solution
$\cot \theta = \dfrac{2xy}{x^2 - y^2} = \dfrac{base}{perp}$
$base = 2xy, perp = x^2 - y^2, hypo = x^2 + y^2$
$\cos \theta \dfrac{b}{h} = \dfrac{2xy}{x^2 + y^2}$