### Instruction: If
\[\tan x = \frac{2ab}{a^2 - b^2},\]where $a > b > 0$ and $0^\circ < x < 90^\circ,$ then find $\sin x$ in terms of $a$ and $b.$.### Response: I know that $\tan x = \frac{\sin x}{\cos x},$ so I can rewrite the given equation as
\[\frac{\sin x}{\cos x} = \frac{2ab}{a^2 - b^2}.\]
I want to isolate $\sin x,$ so I multiply both sides by $\cos x$ to get
\[\sin x = \frac{2ab\cos x}{a^2 - b^2}.\]
Now I need to find $\cos x$ in terms of $a$ and $b.$
I can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to do that.
I substitute the expression for $\sin x$ into the identity and get
\[\left(\frac{2ab\cos x}{a^2 - b^2}\right)^2 + \cos^2 x = 1.\]
I simplify this equation by expanding the square and multiplying by the common denominator $(a^2 - b^2)^2.$
I get
\[4a^2 b^2 \cos^2 x + (a^2 - b^2)^2 \cos^2 x = (a^2 - b^2)^2.\]
I factor out $\cos^2 x$ and rearrange the terms to get
\[\cos^2 x \left(4a^2 b^2 + (a^2 - b^2)^2\right) = (a^2 - b^2)^2.\]
I divide both sides by the factor in parentheses and take the square root to get
\[\cos x = \pm \frac{a^2 - b^2}{\sqrt{4a^2 b^2 + (a^2 - b^2)^2}}.\]
I need to choose the correct sign for $\cos x.$
Since $0^\circ < x < 90^\circ,$ I know that $\cos x > 0.$
So I pick the positive sign and get
\[\cos x = \frac{a^2 - b^2}{\sqrt{4a^2 b^2 + (a^2 - b^2)^2}}.\]
Now I can plug this expression into the formula for $\sin x$ and simplify.
I get
\[\sin x = \frac{2ab\left(\frac{a^2 - b^2}{\sqrt{4a^2 b^2 + (a^2 - b^2)^2}}\right)}{a^2 - b^2}.\]
I cancel out the common factor of $a^2 - b^2$ and get
\[\sin x = \frac{2ab}{\sqrt{4a^2 b^2 + (a^2 - b^2)^2}}.\]