Showing that $\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$ Since I don't have the answer to this one, I want to make sure I've done this correctly.

Show that 
  $$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$

Since 
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$ 
we have 
$$\tan\left(\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)\right) = x$$ 
Applying the Pythagorean theorem, we learn that $b = 1$, so we have $\tan(x/1) = x$.
It may be a bit short, but I really want to be sure I have this right before I continue on. :)
 A: The connection is way more revealing in its simplicty if you just use trigonometry. For $x$ positive take a right-angled triangle with sides $\overline{AB} = 1$ and $\overline{BC} = x$. Then by definition
$$\angle CAB = \arctan x.$$
The hypotenuse measures $\overline{AC} = \sqrt{1+x^2}$, by Pythagorean Theorem. So the same angle can be also defined as
$$\angle CAB = \arcsin \left(\frac{x}{\sqrt{1+x^2}}\right).$$
For negative $x$ take $\overline{BC} = -x$ and recall the odd symmetry of both sine and tangent. As easy as that.
A: $\tan \arcsin \frac{x}{\sqrt{x^2+1}} = \frac{\sin}{\cos} \arcsin \frac{x}{\sqrt{1+x^2}} = \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}}$ and we know that $\cos^2 x+\sin^2 x=1 $ so $\cos x= \sqrt{1-\sin^2 x}$
so $ \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}} = \frac{\frac{x}{\sqrt{x^2+1}}}{\sqrt{1-\sin^2 \arcsin \frac{x}{\sqrt{1+x^2}}}}=\frac{\frac{x}{\sqrt{x^2+1}}}{\sqrt{1-\frac{x^2}{1+x^2}}}  = \frac{x}{\sqrt{1+x^2} \sqrt{\frac{1}{x^2+1}}} = x$
A: Let's consider the definitions of arcsine and arctangent.
Let $\arctan x = \theta$.  Then $\theta$ is the unique angle in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan\theta = x$.
Let 
$$\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) = \varphi$$
Then $\varphi$ is the unique angle in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that 
$$\sin\varphi = \frac{x}{\sqrt{1 + x^2}}$$
We need to show that $\theta = \varphi$.
Observe that 
$$\left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = \frac{x^2}{1 + x^2} < 1$$
for every real number $x$ since 
$$\frac{x^2}{1 + x^2} < 1 \iff x^2 < 1 + x^2 \iff 0 < 1$$
Hence, 
$$-1 < \frac{x}{\sqrt{1 + x^2}} < 1 \implies -\frac{\pi}{2} < \varphi = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) < \frac{\pi}{2}$$
By the Pythagorean identity $\sin^2\varphi + \cos^2\varphi = 1$, 
\begin{align*}
\cos^2\varphi & = 1 - \sin^2\varphi\\
              & = 1 - \left(\frac{x}{\sqrt{1 + x^2}}\right)^2\\
              & = 1 - \frac{x^2}{1 + x^2}\\
              & = \frac{1 + x^2 - x^2}{1 + x^2}\\
              & = \frac{1}{1 + x^2}
\end{align*}
Since $-\frac{\pi}{2} < \varphi < \frac{\pi}{2}$, $\cos\varphi > 0$, so we take the positive square root.  Thus,
$$\cos\varphi = \frac{1}{\sqrt{1 + x^2}}$$
Hence, 
\begin{align*}
\tan\varphi & = \frac{\sin\varphi}{\cos\varphi}\\
            & = \frac{\frac{x}{\sqrt{1 + x^2}}}{\frac{1}{\sqrt{1 + x^2}}}\\
            & = x\\
            & = \tan\theta
\end{align*}
Since $\theta$ is the unique angle in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan\theta = x$, $\tan\varphi = x$, and $-\frac{\pi}{2} < \varphi < \frac{\pi}{2}$, we may conclude that $\theta = \varphi$.
