Show that $T(S^n) \times \mathbb{R}$ is diffeomorphic to $S^n \times \mathbb{R}^{n+1}$
First how $S^n=f^{-1}(1)$ with $f:\mathbb{R}^{n+1} \rightarrow \mathbb{R}$ given by $f(x)=||x||^2$ then for all $x \in S^{n}$, $Tx(S^n)=\{ v \in \mathbb{R}^{n+1} ; (df_x)(v)=0 \}$ and hence is easy to see that $T(S^n)=\{ (x,v) \in S^n \times \mathbb{R}^{n+1} ; \langle x,v \rangle =0 \}$ then $$T(S^n) \times \mathbb{R} =\{(x,v,t) ; x \in S^n, \langle x,v \rangle=0, t \in \mathbb{R} \}$$ And $$S^n \times \mathbb{R}^{n+1} =\{ (x,w); x \in S^n, w \in \mathbb{R}^{n+1}\}$$
then i try to found one diffeomorphism such that maps the last two entries of $T(S^n) \times \mathbb{R}$ in the last entry of $S^n \times \mathbb{R}^{n+1} $ but i now sure how proceded.
Any hint or help i will be very grateful.