= Solution
Regard the space of real $n\times n$ <matrices> as $\mathbb R^{n^2}$, and use the smooth <determinant> map. For invertible $A$, factor $A+tH=A(I+tA^{-1}H)$. Expanding the <determinant> to first order gives
$$
d(\det)_A(H)=\det(A)\operatorname{tr}(A^{-1}H),
$$
since the linear coefficient of $\det(I+tK)$ is the sum of the diagonal entries of $K$. At every $A$ with <determinant> one, taking $H=A$ makes this <derivative> $n\ne0$. Thus one is a <regular value>, and part (ii) makes $\operatorname{SL}(n,\mathbb R)$ a <submanifold> of <dimension> $n^2-1$.
Its <tangent space> is the kernel of this <differential>. At the identity,
$$
\boxed{T_I\operatorname{SL}(n,\mathbb R)=\{H:\operatorname{tr}H=0\}=\mathfrak{sl}(n,\mathbb R).}
$$
Indeed tangent <derivatives> of curves with constant <determinant> lie in that kernel, and the local regular-value coordinates show every kernel <vector> is tangent to the level set.
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