$\sin 1^\circ$ is irrational but how do I prove it in a slick way? And $\tan(1^\circ)$ is .....

If it's a slick proof you want, nothing beats this proof that the only cases where both $r$ and $\cos(r \pi)$ are rational are where $\cos(r \pi)$ is $-1$, $-1/2$, $0$, $1/2$ or $1$.

If $r=m/n$ is rational, $e^{i \pi r}$ and $e^{-i \pi r}$ are roots of $z^{2n} - 1$, so they are algebraic integers. Therefore $2 \cos(r \pi) = e^{i \pi r} + e^{-i \pi r}$ is an algebraic integer. But the only algebraic integers that are rational numbers are the ordinary integers. So $2 \cos(r \pi)$ must be an integer, and of course the only integers in the interval $[-2,2]$ are $-2,-1,0,1,2$.

$\sin(1^\circ) = \cos(89^\circ)$, and since 89 is relatively prime to 360, the proof for $\cos 1^\circ$ works with almost no change.

More precisely: Assume that $\cos(89^\circ)$ is rational. Then, by the same induction as before with every $1^\circ$ replaced by $89^\circ$ we get that $\cos(89n^\circ)$ is rational for every $n\in\mathbb N$. In particular, since $150\times 89=37\times 360+30$, we get that $$\cos(150\times 89^\circ)=\cos(37\times 360^\circ+30^\circ)=\cos(30^\circ)$$ is rational, a contradiction.

For $\tan(1^\circ)$, a slight variant of the same proof works. Assume that $\alpha = \tan(1^\circ)$ is rational. Then $1+\alpha i$ is in $\mathbb Q[i]$, and then $\tan(n^\circ)$, being the ratio between the imaginary and real parts of $(1+\alpha i)^n$ is also rational. But $\tan(30^\circ)$ is not rational, so $\tan(1^\circ)$ cannot be either.

You can prove it exactly the same way:

Assume by contradiction that $\sin(1^\circ)$ is rational.

Then

$$\cos(2^\circ)=1- 2\sin^2(1^\circ) \mbox{is rational}\,,$$

Now you can also prove that $\cos 4^\circ$ is rational.

Using

$$\cos((2n+2)^\circ)+\cos((2n-2) ^\circ)=2\cos(2^\circ)\cdot \cos((2n)^\circ) \,,$$

you can prove by induction that $\cos(2n^\circ)$ is rational, and you get your contradiction...

Added

If $\tan(1^\circ)$ is rational, then

$$\cos(2^\circ) =\frac{1- \tan^2(1^\circ)}{1+\tan^2(1^\circ)}$$ is also rational...

Alternately, if you are not familiar with this relation, note that

$$\cos^2(1^\circ)= \frac{1}{\sec^2(1^\circ)}= \frac{1}{1+ \tan^2(1^\circ)}$$ is rational, thus

$$\cos(2^\circ)=2\cos^2(1^\circ)-1$$ is rational.

The first part of the proof finishes this part too...

P.S. You can actually prove by induction the following result: if $\cos(x)$ is rational, then $\cos(nx)$ is also rational.