Hitting-time Brownian Bridge

Question

Let $(B_t)_{t\in[0,1]}$ be a standard Brownian bridge, that is $B_0=B_1=0$. I'm interested in $\tau_a:=\inf\{t\in[0,1],B_t=a\}\in[0,1]\cup\{+\infty\}$ for $a>0$. One of the choices for $(B_t)_{t\in[0,1]}$ is $B_t=(1-t)W_{\frac{t}{1-t}}$ where $(W_t)_{t\geqslant 0}$ is a standard brownian motion. Thus for $a>0$, $$ B_t=a\iff W_{\frac{t}{1-t}}=\frac{a}{1-t}=a\left(1+\frac{t}{1-t}\right) $$ Since $t\mapsto\frac{t}{1-t}$ is inscreasing, the "first" $t$ such that $B_t=a$ is such that $\frac{t}{1-t}=T_{a,a}$ where $T_{a,b}:=\inf\{t\geqslant 0,W_t=a+bt\}$. It is known that $T_{a,b}<+\infty$ a.s, therefore $\tau_a=\frac{T_{a,a}}{1+T_{a,a}}$ and $\tau_a<+\infty$ a.s, which means that $(B_t)_{t\in[0,1]}$ hits all the $a>0$ in finite time. This is not true at all but I can't spot the mistake.