I want to show that
(a) $TO(m) = \{(M,MA) \mid M \in O(m), A \in T_{Id}O(m)\}$
and
(b) $TSL(m) = \{(M,MA) \mid M \in SL(m), A \in T_{Id}SL(m)\}.$
By definition, $TX = \{(x,y) \mid x \in X, y \in T_xX\}$.
For (a), $$T_{M}O(m) = \{H \mid H^tM + HM^t = 0\}.$$ I can show that $\{MA \mid M \in O(m), A \in T_{Id}O(m)\} \subset T_MO(m)$. Is possible to show that this sets are equal? If yes, then the problem is solved.
For (b), $$T_{Id}SL(m) = \{A \mid tr{A} = 0\},$$ but I could not to figure out $T_M SL(m)$. My idea is also to show that $\{MA \mid M \in SL(m), A \in T_{Id}SL(m)\}$.
I'm beginner in Differential Geometry. So, if there are better ways to do that, I would like to know
Yes. For any of these matrix groups $G$, $T_M G = \{ MA \mid A \in T_I G\}$. That's why we study carefully the tangent space to the identity. It's called the Lie algebra of $G$. It “sees” a lot of the properties of $G$, but has the advantage of being a vector space.