$\sin(\tan^{-1}(x/a)) = \frac{x/a}{\sqrt{1+(x/a)^2}}$ Why does:
$\sin(\tan^{-1}(x/a)) = \frac{x/a}{\sqrt{1+(x/a)^2}}$

 A: Quick answer: consider a right angle triangle with opposite side $x$ and adjacent side $a$.
Longer answer:
$$\begin{align*}
\sin\left(\tan^{-1}\frac xa\right)
&= \frac{\tan\left(\tan^{-1}\frac xa\right)}{\sec\left(\tan^{-1}\frac xa\right)}\\
&= \frac{\tan\left(\tan^{-1}\frac xa\right)}{\sqrt{1+\tan^2\left(\tan^{-1}\frac xa\right)}}\\
\end{align*}$$
The square root is positive because $\tan^{-1} x \in (-\frac\pi2, \frac \pi 2)$, and $\sec$ is positive in that interval.
A: Draw a right angle triangle.

Using SOH-CAH-TOA first diagram comes from $\tan(\theta)=\frac{\text{opp}}{\text{adj}}=\frac{x/a}{1}$, so $\displaystyle \theta=\tan^{-1}\left(\frac{x}{a}\right)$
Then, using $\sin(\theta)=\frac{\text{opp}}{\text{hyp}}$ and the Pythagorean Theorem for the hypotenuse of the triangle, we get $\displaystyle \sin(\theta)=\sin\left(\tan^{-1}\left(\frac{x}{a}\right)\right)=\frac{x/a}{\sqrt{1+\frac{x^2}{a^2}}}$.
A: Let $\tan^{-1}\dfrac xa=y,-\dfrac\pi2<y<\dfrac\pi2\implies\cos y>0$
$\tan y=\dfrac xa,$
$$\dfrac{\sin y}{\dfrac xa}=\dfrac{\cos y}1=\pm\sqrt{\dfrac{\sin^2y+\cos^2y}{\left(\dfrac xa\right)^2+1}}$$
$\implies\cos y=\dfrac1{\sqrt{\left(\dfrac xa\right)^2+1}}$
