$\mathbb CP^n$ with a quadric removed is homeomorphic to $T(\mathbb RP^n)$. Let $V_f$ be the zero set of a quadratic $z_1^2+\dots +z_{n}^2$ in $\mathbb CP^{n}$. I would like to show that 
$P^{n}(\mathbb C) \setminus V_f$ is diffeomorphic to the total space of the tangent bundle $T (\mathbb RP^{n})$.

Some of my observations: 
Let $\tilde{V_f}$ be the preimage of $V_f$ under the projection $S^{2n+1} \to \mathbb CP^n$. Rewriting the condition that
letting $\mathbf z \in \mathbb C^{n+1}$ be a vector $z_n^2+\dots +z_0^2=0$ in real terms, we get that $(|x|^2-|y|^2)+2i \langle x, y \rangle=0$, so $|x|^2=|y|^2$ and the inner product is zero.
However, since $|x|^2+|y|^2=1$ if it is to be on $S^{2n+1}$, we get that $|x|$ and $|y|$ are both $\frac{1}{\sqrt{2}}$.
Hence, a diffeomorphism $\tilde{V_f} \to V_2(\mathbb R^{n+1})$ (where the codomain is the set of orthonormal $2$-frames in $\mathbb R^{n+1}$) is given by $(x,y) \mapsto (\sqrt{2}x,\sqrt{2}y)$.
From this, we can conclude that $V_f \cong G_2(\mathbb R^{n+1})$.
Is there a way to conclude from here?
 A: Let $\pi:S^{2n+1}\rightarrow \mathbb{C}P^n$ be the canonical projection and let $\tilde V_f=\pi^{-1}(V_f)\subseteq S^{2n+1}$ be the inverse image of $V_f\subseteq \mathbb{C}P^n$. As you have shown, $\tilde V_f\cong V_2(\mathbb{R}^{n+1})$, and $\pi$ restricted to this space is an $S^1$-principal fibration over $V_f\cong Gr_2(\mathbb{R}^{n+1})$. Let
$$F=\{(z_0,\dots,z_n)\in S^{2n+1}\setminus\tilde V_f\mid z_0^2+\dots+z_n^2=1\}$$
be the Milnor fibre of the polynomial $f$. Notice that $\pi|_F:F\rightarrow \mathbb{C}P^n\setminus V_f$ is surjective. 
Writing $\underline z=(z_0,\dots,z_n)\in S^{2n+1}\subseteq \mathbb{C}^{n+1}$ as its real and imaginary parts $\underline z=\underline x+i\underline y$, we define a map
$$\tilde \Psi:F\rightarrow TS^n,\qquad \underline x+i\underline y\mapsto\left(\frac{\underline x}{|\underline x|},\underline y\right),$$
where we have identified the tangent bundle $TS^n=\{(\underline x,\underline y)\in S^{n}\times \mathbb{R}^{n+1}\mid \underline x\cdot \underline y=0\}$. We see that this map is well-defined using the calculation you have already supplied. In fact, writing
$$F=\{(\underline x,\underline y)\in\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}\mid \underline x^2=\underline y^2+1,\;\underline x\cdot \underline y=0\}$$
we see quite easily that $\tilde \Psi$ is one-to-one and onto. It is clearly smooth and its smooth inverse is easy to write down, so it is seen to be a diffeomorphism.
We observed before that $\pi|_F:F\rightarrow \mathbb{C}P^n\setminus V_f$ is surjective, although it is not difficult to see that it fails to be $S^1$-principal. What is true, however, is that if we consider the $S^1$ orbit in $S^{2n+1}\setminus\tilde V_f$ of a point $\underline z\in F$, then we find that
$$(S^1\cdot \underline z)\cap F=\{\pm\underline z\}$$
consists of exactly two points. In fact $\pi^{-1}(\pi(\underline z))\cap F=\{\pm\underline z\}$.
The point is that $F$ is closed under the restriction of the $S^1$-action to its $\mathbb{Z}_2$-subgroup and we have a principal fibration
$$\mathbb{Z}_2\hookrightarrow F\xrightarrow{\pi|} \mathbb{C}P^n\setminus V_f.$$
Now recall that map $\tilde \psi:F\rightarrow TS^n$. it is clear from its definition that this map is $\mathbb{Z}_2$-equivariant with respect to the natural $\mathbb{Z}_2$-action on $TS^n$ induced by the tangent map of the antipodal map on $S^n$. Thus there is an induced map of $\mathbb{Z}_2$-orbit spaces
$$\Psi:\mathbb{C}P^n\setminus V_f\rightarrow T\mathbb{R}P^n\cong (TS^n)/\mathbb{Z}_2.$$
Since the fibre-preserving map $\tilde\Psi$ is a diffeomorphism, so too is its quotient $\Psi$, which is therefore the map you have been looking for.
