rewrite the expression as an algebraic expression in x. \n( sin ( \tan ^ { - 1 } ( x ) ) )

Explanation:

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

By the definition of the inverse tangent function, $\tan\theta=x$, where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$. And we know that $\tan\theta=\frac{x}{1}$, so we can consider a right - triangle where the opposite side is $x$ and the adjacent side is $1$.

Step2: Find the hypotenuse

Using the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$, where $a = 1$ and $b=x$. The hypotenuse $r=\sqrt{1 + x^{2}}$.

Step3: Find $\sin\theta$

Since $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\theta=\tan^{-1}(x)$, then $\sin(\tan^{-1}(x))=\sin\theta$. Substituting the values from the right - triangle, we get $\sin\theta=\frac{x}{\sqrt{1 + x^{2}}}$.

Answer:

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