= Solution
For $X=e^\beta\cos\theta$ and $Y=e^\beta\sin\theta$, apply the two-variable <Itô formula>. Independence gives $[\beta,\theta]=0$, while each individual bracket is $t$. The second derivatives in the two variables cancel, yielding
$$
\boxed{dX_t=X_t\,d\beta_t-Y_t\,d\theta_t,\qquad dY_t=Y_t\,d\beta_t+X_t\,d\theta_t.}
$$
Their integrands are continuous adapted and locally bounded, so both coordinates are <continuous local martingales>. The <quadratic variation of a stochastic integral> gives
$$
\boxed{[X]_t=[Y]_t=A_t:=\int_0^t(X_s^2+Y_s^2)ds=\int_0^te^{2\beta_s}ds,\qquad[X,Y]_t=\int_0^t(X_sY_s-Y_sX_s)ds=0.}
$$
Here is a full proof that the clock diverges. Let $S_0=0$, and successively define
$$
U_j=\inf\{t\geq S_j:|\beta_t|=1\},\qquad S_{j+1}=\inf\{t\geq U_j:\beta_t=0\}.
$$
By <recurrence of one-dimensional Brownian motion>, all these times are finite almost surely. By the <Strong Markov property> at the returns $S_j$, the variables $D_j=(U_j-S_j)\wedge1$ are independent identically distributed, lie in $(0,1]$ almost surely, and have a strictly positive mean. The <strong law of large numbers> gives $\sum_jD_j=\infty$ almost surely. On $[S_j,U_j]$, $\beta_s\geq-1$, so
$$
\int_0^\infty e^{2\beta_s}ds\geq\sum_j\int_{S_j}^{U_j}e^{2\beta_s}ds\geq e^{-2}\sum_jD_j=\infty.
$$
The segments are disjoint, and their divergent duration also ensures they cannot accumulate at a finite time. This proves <Divergence of a driftless Brownian exponential clock> and
$$
\boxed{[X]_\infty=[Y]_\infty=\infty\quad\text{almost surely}.}
$$
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