$\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.

$$\begin{aligned} \cos( \tan ( \alpha)) = \sin( \alpha) &\\ t \rightarrow \frac{\pi}{2}- \alpha &\\ \cos( \tan ( \frac{\pi}{2} -t)) = \cos(t) &\\ \tan( \frac{\pi}{2} -t) = t &\\ \cot(t) = t &\\\end{aligned}$$
Which does not have an analytical solution
