$\cos (\tan \alpha)=\sin \alpha $ I can't find where is my mistake so if someone can enlighten me...
$\tan \alpha=\arccos (\sin \alpha) $
$\tan \alpha=\arccos (\cos (\alpha -\frac{\pi }{2}))= \alpha -\frac{\pi }{2}$
$\frac{d}{d\alpha }\tan \alpha=\frac{d}{d\alpha }(\alpha -\frac{\pi }{2})  $
$\frac{1}{\cos (\alpha)^{2}}=1  $
$\alpha =\pi n$
there is something wrong and I don't really know.

The fact that $\alpha$ satisfies $F(\alpha)=G(\alpha)$ does not imply that $F'(\alpha)=G'(\alpha)$. For instance, $x=0$ is a solution to $x=x^2$, but not a solution to $1=2x$.
Added: There are of course a few other issues of purely algebraic nature, namely that $\cos t=c$ does not imply $t=\arccos c$, and that $\arccos \cos c\ne c$.
