1972 AMC 12 Question 20
AMC12 · 1972 AMC 12
If $\tan x=\dfrac{2ab}{a^2-b^2},$ where $a\gt b\gt0$ and $0^\circ\lt x\lt90^\circ,$ then $\sin x$ is equal to:
$\dfrac ab$
$\dfrac ba$
$\dfrac{\sqrt{a^2-b^2}}{2a}$
$\dfrac{\sqrt{a^2-b^2}}{2ab}$
$\dfrac{2ab}{a^2+b^2}$