In fact the subgroup generated by $SO(n)$ and $tSO(n)t^{-1}$ is certainly connected by arc (as the union of images of arc-connected spaces $SO(n)^k, k\in \bf N$ by the obvious product map) So it is an immersed Lie subgroup (Yamabe). But the argument of Venkataraman shows that there is no Lie algebra between $so(n)$ and $sl(n)$, and the Lie algebra of this immersed Lie subgroup is $sl(n)$. By connectedness this group is $SL(n)$.

In fact the subgroup generated by $SO(n)$ and $tSO(n)t^{-1}$ is certainly connected by arc (as the union of images of arc-connected spaces $SO(n)^k, k\in \bf N$ by the obvious product map) So it is an immersed Lie subgroup (Yamabe). But the argument of Venkataramana shows that there is no Lie algebra between $so(n)$ and $sl(n)$, and the Lie algebra of this immersed Lie subgroup is $sl(n)$. By connectedness this group is $SL(n)$.

In fact the subgroup generated by $SO(n)$ and $tSO(n)t^{-1}$ is certainly connected by arc (as the union of images of arc-connected spaces $SO(n)^k, k\in \bf N$ by the obvious product map) So it is an immersed Lie subgroup (Yamabe). But the argument of Venkataraman shows that there is no a Lie algebra between $so(n)$ and $sl(n)$, and the Lie algebra of this immersed Lie subgroup is $sl(n)$. By connectedness this group is $SL(n)$.

In fact the subgroup generated by $SO(n)$ and $tSO(n)t^{-1}$ is certainly connected by arc (as the union of images of arc-connected spaces $SO(n)^k, k\in \bf N$ by the obvious product map) So it is an immersed Lie subgroup (Yamabe). But the argument of Venkataraman shows that there is no Lie algebra between $so(n)$ and $sl(n)$, and the Lie algebra of this immersed Lie subgroup is $sl(n)$. By connectedness this group is $SL(n)$.