I know that for $z \in \mathbb{C}^{n+1} \setminus \left\{ 0 \right\} $, denoted by $L_z$ the complex line in $\mathbb{C}^{n+1}$ passing through $z$ and the origin, i.e., $$L_z = \left\{ \lambda z \in \mathbb{C}^{n+1} \space\ \big| \space\ \lambda \in \mathbb{C} \right\}$$ Note that for a point $[z] \in \mathbb{CP}^n$, as a set $[z] = L_z \setminus \left\{ 0 \right\}$. The tautological line bundle $\pi : T \to \mathbb{CP}^n$ I know is given by:
a) The total space $T = \left\{ ([z] , w ) \in \mathbb{CP}^n \times \mathbb{C}^{n+1} \space\ \big| \space\ w \in L_z \right\}$ and
b) $\pi : T \to \mathbb{CP}^{n}$ is given by the projection onto the first factor.
So if we let $U_i = \left\{ [z] = [z_1 : \dots : z_{n+1} ] \in \mathbb{CP}^{n} \space\ \big| \space\ z_i \neq 0 \right\} \hspace{.5cm} i = 1, \dots , n+1.$
I'm not sure how I would go about finding local trivializations $$g_i : \pi^{-1}(U_i) \to U_i \times \mathbb{C} , \space\ 1 \leq i \leq n+1,$$ for the bundle $T$ or finding transition functions $$g_{ji} : U_i \cap U_j \to GL(1,\mathbb{C}), 1 \leq i \neq j \leq n+1,$$ for the bundle $T$.
Any help would be appreciated.
There is a section $\sigma_i : U_i \to \pi^{-1}(U_i)$ given by
$$\sigma_i([z_1: \dots : z_{n+1}]) = ([z_1: \dots : z_{n+1}], (z_1', \dots, z_{n+1}'))$$
where $z_j' = z_jz_i^{-1}$. As $z_i' = z_iz_i^{-1} = 1 \neq 0$, the section $\sigma_i$ is nowhere-zero. For every $w \in L_{[z]}$, there is a unique $\lambda \in \mathbb{C}$ such that $w = \lambda\sigma_i([z])$. This gives rise to the local trivialisation $g_i : \pi^{-1}(U_i) \to U_i\times\mathbb{C}$, $([z], w) \mapsto ([z], \lambda)$.
Note that $g_i^{-1}([z], \lambda) = ([z], \lambda\sigma_i([z]))$ so $g_j(g_i^{-1}([z], \lambda)) = ([z], \lambda')$ where $\lambda'\sigma_j([z]) = \lambda\sigma_i([z])$. It follows that $\lambda' z_kz_j^{-1} = \lambda z_kz_i^{-1}$ and hence $\lambda' = z_jz_i^{-1}\lambda$. As
$$([z], z_jz_i^{-1}\lambda) = ([z], \lambda') = g_j(g_i^{-1}([z], \lambda)) = ([z], g_{ji}([z])\lambda),$$
we see that the transition functions for this line bundle are $g_{ji}([z]) = z_jz_i^{-1} = \dfrac{z_j}{z_i}$.