Showing the tangent space of $SL(n,\mathbb{C})=\{A\in GL(n,\mathbb{C}):\det(A)=1\}$ is $S=\{A\in GL(n,\mathbb{C}):\text{trace(A)}=0\}$
I have the definition of tangent space at a matrix $A$ is $T_A(M)=\{X\in M: \exists\gamma:[0,1]\to M,\text{is differentiable}, \gamma'(0)=X,\gamma(0)=A$
So if I take some matrix $B\in T_A(SL(n,C))$ I want to show it has $0$ trace , to show the tangent is contained in my set $S$
But I don't really have anything linking trace of a matrix to its determinant except for $\det(e^A)=e^{Tr(A)}$
The map $X\mapsto \exp X$ is a diffeomorphism from a neigborhood of $0_n$ to a neighborhood of $I_n$. Let $\gamma \colon [0,1[ \to SL(n,\mathbb{C})$, $\gamma(0)= I_n$. For some $\epsilon>0$ we can write $\gamma(t) = \exp \eta(t)$, with $\eta(\cdot)$ smooth, and $\eta(0) = O_n$. Now $\gamma(t) \in SL(n,\mathbb{C})$ implies $Tr \,\eta(t) = 0$ for all $t \in [0, \epsilon)$. We have $\gamma'(t) = \exp (\eta(t)) \cdot \eta'(t)$, and $\gamma'(0) = \eta'(0)$ has also trace $0$.
Therefore, the tangent space of $SL(n, \mathbb{C})$ at $I_n$ is $sl(n, \mathbb{C})$, the subspace of $n\times n$ complex matrices with trace $0$.
Note: Here we determined the tangent space of $SL(n,\mathbb{C})$ at $I_n$. At an arbitrary element $A\in SL(n, \mathbb{C})$ it will be $A\cdot sl(n, \mathbb{C})= sl(n,\mathbb{C})\cdot A$.