Solution 1:

Let $Q\subseteq\mathbf C^{n+1}$ be the affine quadric defined by the equation $\sum z_i^2=1$. The map $$ f\colon TS^n\rightarrow Q $$ defined by $$ z=f(x,y)=x\sqrt{1+||y||^2}+y\sqrt{-1} $$ does the job, where $||y||^2=\sum y_i^2$. Indeed, one has $f(x,y)\in Q$ since $$ \sum_{i=0}^n z_i^2=\sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_i\sqrt{-1}\sqrt{1+||y||^2}=\\ 1+||y||^2-||y||^2+2\sqrt{-1}\sqrt{1+||y||^2}\sum x_iy_i=1, $$ for $(x,y)\in TS^n$.

The map $f$ is a diffeomorphism since its inverse is $$ g\colon Q\rightarrow TS^n $$ defined by $$ g(z)=\left(\frac{x}{\sqrt{1+||y||^2}}, y\right), $$ where $z=x+y\sqrt{-1}$. One has $g(z)\in TS^n$ since $$ ||x||^2-||y||^2=1 $$ and $$ 2\sqrt{-1}\sum x_iy_i=0 $$ for $z\in Q$.