My textbook has this expression, but I can't figure (either geometrically or analytically) why is this true and how do I get from one expression to the other.
$$\sin\left[\arctan\left(\frac{x}{a}\right)\right] = \frac{x}{a\sqrt{\frac{x^2}{a^2} + 1}}$$
I'll show you how to work out anything in trig(inversetrig) form, I'll do $\cos{\tan^{-1}{\frac{x}{a}}}$, and you can figure out your case with much the same method.
Let $\alpha=\tan^{-1}{\frac{x}{a}}$
$\tan{\alpha}=\frac{x}{a}$
Now you know your trig ratios and tan is opposite on adjacent which means your triangle has sides $x$ (opposite), $a$ (adjacent) and $\sqrt{x^2+a^2}$ (hypotenuse, use pythagoras to figure it out)
Therefore $\cos{\alpha}=\frac{a}{\sqrt{x^2+a^2}}=\frac{a}{a\sqrt{\frac{x^2}{a^2}+1}}$ (remember $\cos{\alpha}=\cos{\tan^{-1}{\frac{x}{a}}}$).
Now see if you can apply this method to $\sin{\tan^{-1}{\frac{x}{a}}}$ :D
Draw a right triangle and label one of the acute angles $\theta$. Now, label the side opposite $\theta$ as 'x' and the side adjacent to $\theta$ as 'a'. From the pythagorean theorem, the hypotenuse is then $\sqrt{x^2+a^2}$. We then see directly that $sin(\theta) = \frac{x}{\sqrt{x^2+a^2}}$.
Now, let's express $\theta$ using the arctangent function. Since $\tan(\theta) = x/a$, it follows that $\theta = \arctan(\frac{x}{a})$. If we substitute $\theta$ into the sine function, we have the result you're looking for:
$$\sin\left(\arctan\left(\frac{x}{a}\right)\right) = \frac{x}{\sqrt{x^2+a^2}}.$$
All that's left is to factor an $a^2$ from the argument of the square root and you're there. $$\sin\left(\arctan\left(\frac{x}{a}\right)\right) = \frac{x}{a\sqrt{\frac{x^2}{a^2}+1}}.$$
If we assume that $\dfrac xa=\tan\theta$, the identity $$\sin\left(\arctan\left(\frac xa\right)\right)=\frac{\dfrac xa}{\sqrt{\left(\dfrac xa\right)^2+1}}$$
becomes
$$\sin(\theta)=\frac{\tan\theta}{\sqrt{\tan^2\theta+1}}=\frac{\sin\theta}{\sqrt{\sin^2\theta+\cos^2\theta}}.$$
[Note that the signs are correct because in the range of the arc tangent, the cosine is positive.]
