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. :)

$\tan \arcsin \frac{x}{\sqrt{x^2+1}} = \frac{\sin}{\cos} \arcsin \frac{x}{\sqrt{1+x^2}} = \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}}$ 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$
