= Solution
The tangent matrices are
$$
X=H'(0)=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad Y=g'(0)=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.
$$
Use the derivative of conjugation from part (i), rather than assuming antisymmetry. Since $H(s)^{-1}=\begin{pmatrix}1&-s\\0&1\end{pmatrix}$, direct multiplication gives
$$
\operatorname{Ad}_{H(s)}Y=H(s)YH(s)^{-1}=\begin{pmatrix}-s&1+s^2\\-1&s\end{pmatrix}.
$$
Differentiating at zero yields
$$
\boxed{[X,Y]=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.}
$$
For the reverse order put $c=\cos t$, $s=\sin t$. The other conjugation is
$$
\operatorname{Ad}_{g(t)}X=\begin{pmatrix}cs&c^2\\-s^2&-sc\end{pmatrix}.
$$
Its derivative at zero is
$$
\boxed{[Y,X]=\begin{pmatrix}1&0\\0&-1\end{pmatrix}=-[X,Y].}
$$
Both computations use actual curves in the <special linear group>; the opposite off-diagonal signs in $g(t)$ ensure its determinant is one.
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