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 $\arccos \cos c\ne c$.
The function $$f(\alpha)=\cos (\tan (\alpha))-\sin (\alpha)$$ will not show explicit roots (remember that is already the case of $x=\cos(x)$). SO, you will need a numerical method, so a reasoble guess for the solution you are looking for since there is an infinity of roots.
If you look for the ones close to zero, compose Taylor series to get $$f(\alpha)=1-\alpha -\frac{\alpha ^2}{2}+O\left(\alpha ^3\right)$$ Solving the quadratic will give two roots $$a_-=-1-\sqrt{3} \sim -2.732 \qquad \text{and} \qquad a_+=-1+\sqrt{3}\sim 0.732$$ while the "exact" solutions (given by Newton method) are $$a_-\sim -1.287 \qquad \text{and} \qquad a_+\sim 0.710$$
Edit
Staring from @logichtech's answer, we need to find the zero's of function $$f(t)= \cot(t) -t$$ which is not very well conditioned because of the dicontinuities induced by the cotangent.
It is then much better to consider function $$g(t)=\cos(t)-t \sin(t)$$ the roots of which being closer and closer to $n\pi$ because of the sine.
So, let expand as a Taylor series function $g(t)$ around $t=n\pi$. This will give $$g(t)=\sum_{k=0}^\infty a_k (t-n\pi)^k$$ where the coefficients make the sequance $$\left\{1,-\pi ,-\frac{3}{2},\frac{\pi }{2},\frac{5}{24},-\frac{\pi }{24},-\frac{7}{720},\frac{\pi }{720},\frac{1}{4480},-\frac{\pi }{40320},-\frac{11}{3628800}\right\}$$ Now, using series reversion $$t=k+\frac{1}{k}-\frac{4}{3 k^3}+\frac{53}{15 k^5}-\frac{1226}{105 k^7}+\frac{13597}{315 k^9}-\frac{1531127}{8960 k^{11}}+O\left(\frac{1}{k^{12}}\right)$$ where $k=n \pi$.
Using the above, compare the approximation and the solution given by Newton method for the $n^{th}$ root $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 2 & 3.42544789218750 & 3.42561845948170 \\ 3 & 6.43729815226868 & 6.43729817917194 \\ 4 & 9.52933440521558 & 9.52933440536196 \\ 5 & 12.6452872238531 & 12.6452872238566 \\ 6 & 15.7712848748157 & 15.7712848748159 \\ 7 & 18.9024099568600 & 18.9024099568600 \end{array} \right)$$
For the very first root, the same procedure would lead to $$t=\frac{200683117717}{95234227200 \sqrt{6}}\sim 0.860285$$ while the "exact" value is $0.860334$.
$$\begin{aligned} \cos( \tan ( \alpha)) = \sin( \alpha) &\\ t \rightarrow \frac{\pi}{2}- \alpha &\\ \cos( \tan ( \frac{\pi}{2} -t)) = \cos(t) &\\ \tan( \frac{\pi}{2} -t) = t &\\ \cot(t) = t &\\\end{aligned}$$ Which does not have an analytical solution
