If tan x=(2ab)/(a^2-b^2), where a> b>0 and 0°< x<90°, then sin x is equal to:

1972 AMC 12 Problem 20

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20.

\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}

Model the tangent as the ratio of legs

2ab

and

a^2-b^2

The corresponding hypotenuse simplifies because

(2ab)^2+(a^2-b^2)^2=(a^2+b^2)^2

Use a right triangle with opposite leg

2ab

and adjacent leg

a^2-b^2.

Its hypotenuse is

\begin{aligned} &\sqrt{(2ab)^2+(a^2-b^2)^2}\\ &\qquad=\sqrt{(a^2+b^2)^2}\\ &\qquad=a^2+b^2. \end{aligned}

Hence

\sin x=\frac{2ab}{a^2+b^2}.

Therefore, the correct answer is E.