Simplifying the identity: $\sin(\arctan\frac x a)$

I want to simplify the identity: $\sin(\arctan\frac x a)$. I can do this with geometry. In a triangle $ABC$, let $AB=x$, $BC=a$ and $\angle B=90^\circ$. Then, $\arctan\frac xa=\angle C$ and $\sin(\arctan\frac x a)=\frac{AB}{AC}=\frac{x}{\sqrt{a^2+x^2}}$. But how to simplify this without geometry i.e using trigonometric formulas and algebra?

I tried using the formula: $\sin x=\frac{2\tan\frac x2}{1+\tan^2\frac x2}$ and got the identity $$\sin(\arctan\frac x a)=\frac{2\tan(\frac {\arctan\frac xa}2)}{1+\tan^2 (\frac {\arctan\frac xa}2)}$$ But this doesn't seem to work.

Draw a right-angled triangle with the non-hypotenuse edges having length x and a. The angle opposite x is arctan of x/a. Now, you want to compute sine of this angle. You should know what to do!

Easier with a trig drawing that is the basis of the Pythagorean triangle/thm.

Let $ t= \dfrac{x}{a}= \dfrac{t}{1}=\tan \theta $

Read off directly from the drawing

$$\sin \theta =\pm \frac{ t}{\sqrt{ 1+t^2}} $$

Due to the radical sign we needed to insert $\pm$in front of the fraction.

This ambiguity meant that our $\theta $ was either in the first or in the third quadrants.