Constructable Trigonometric Inverses By doing some right triangle gymnastics, we can derive things like
$\cos(\arctan x) = \frac{1}{\sqrt{1+x^2}}$, for $x>0$
$\cos(\arcsin x) = \sqrt{1-x^2}$
$\tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}}$
What about $\arctan\cos(x)$, $\arcsin(\tan x)$, etc? I understand that in this case $x$ is treated as an angle, not a ratio of side lengths and that it is impossible to construct the same kind of right triangle relations for these formulas. However, is there a particularly compelling non-geometric reason why the reverse application of these functions is intractable?
 A: The identities you list can all be derived by expressing the trigonometric functions in terms of each other, e.g. (glossing over sign issues)
$$\tan x=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-\sin^2x}}\;,$$
and thus
$$\tan(\arcsin x)=\frac{\sin (\arcsin x)}{\sqrt{1-\sin^2(\arcsin x)}}=\frac{x}{\sqrt{1-x^2}}\;.$$
So these identities hold because there are identities between the trigonometric functions, which in a sense are due to Euler's formula.
You can get similar identities in the other direction for $\arcsin$ and $\arccos$ because there's a suitable relationship between these two:
$$\arccos x=\frac\pi2-\arcsin x\;,$$
and thus
$$\arccos(\sin x)=\frac\pi2-\arcsin (\sin x)=\frac\pi2-x\;.$$
That you can't get them for $\arctan\cos(x)$ and $\arcsin(\tan x)$ has to do with the fact that the identities between the corresponding inverse trigonometric functions have different arguments. For instance,
$$\arctan x=\arcsin\frac x{\sqrt{x^2+1}}\;,$$
but you can't turn this into a formula for $\arctan(\sin x)$ because the argument on the right-hand side isn't $x$.
