Let $X = \mathbb P^n$ be the usual projective space over an algebraically closed field. For what values of $d \in \mathbb Z$ does the twisted tangent bundle $E = TX \otimes \mathcal O_X(d-1)$ have a globally nonvanishing section?
My attempt to solve this problem was to consider the twisted Euler sequence
$$0 \to \mathcal O(d-1) \to \mathcal O(d)^{n+1} \to E \to 0$$
Then $E$'s total Chern class is
$$ c(E) = \dfrac {c(\mathcal O(d))^{n+1}} {c(\mathcal O(d-1))} = \dfrac {(1 + dH)^{n+1}} {1 - (1-d)H} = \sum_{j=0}^n \binom {n+1} j d^j H^j \cdot \sum_{k=0}^n (1-d)^k H^k $$
Extracting the top Chern class, we have
$$c_n(E) = aH^n = \sum_{j=0}^n \binom {n+1} j d^j (1 - d)^{n-j} H^n$$
Hence the coefficient is
$$ a(1-d) = \sum_{j=0}^n \binom {n+1} j d^j (1 - d)^{n+1-j} = \left[ \sum_{j=0}^{n+1} \binom {n+1} j d^j (1 - d)^{n+1-j} \right] - d^{n+1} $$
$$a = \dfrac {1 - d^{n+1}} {1 - d} = 1 + d + d^2 + \dots + d^n$$
The only way to solve $a = 0$ while keeping $d$ an integer is to set $d = -1$ and $n$ odd. Does $TX \otimes \mathcal O_X(-2)$ actually have a globally nonvanishing section for $n$ odd?
On $\mathbb{P}^1$, $T_{\mathbb{P}^1}(d-1) = \mathcal{O}_{\mathbb{P}^1}(d-3)$, so it has a nowhere-zero section precisely when $d = 3$. For $n \geq 2$, $T_{\mathbb{P}^n}(-2)$ has no global sections, hence from your computations, there does not exist a nowhere-zero section of any twists of $T_{\mathbb{P}^n}$.