find the exact value of the expression in terms of x: sin(2 tan^(-1)(x)) =

Explanation:

Step1: Let $\theta=\tan^{- 1}(x)$

So, $\tan\theta = x=\frac{x}{1}$, and we can consider a right - triangle where the opposite side to angle $\theta$ is $x$ and the adjacent side is $1$. By the Pythagorean theorem, the hypotenuse $r=\sqrt{1 + x^{2}}$. Then $\sin\theta=\frac{x}{\sqrt{1 + x^{2}}}$ and $\cos\theta=\frac{1}{\sqrt{1 + x^{2}}}$.

Step2: Use the double - angle formula for sine

The double - angle formula for sine is $\sin(2\alpha)=2\sin\alpha\cos\alpha$. Here, $\alpha = \tan^{-1}(x)=\theta$, so $\sin(2\tan^{-1}(x))=2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))$. Substituting $\sin\theta=\frac{x}{\sqrt{1 + x^{2}}}$ and $\cos\theta=\frac{1}{\sqrt{1 + x^{2}}}$ into the double - angle formula, we get $\sin(2\tan^{-1}(x)) = 2\times\frac{x}{\sqrt{1 + x^{2}}}\times\frac{1}{\sqrt{1 + x^{2}}}$.

Step3: Simplify the expression

$2\times\frac{x}{\sqrt{1 + x^{2}}}\times\frac{1}{\sqrt{1 + x^{2}}}=\frac{2x}{1 + x^{2}}$.

Answer:

$\frac{2x}{1 + x^{2}}$