= Solution
Put $A_t=e^{-W_t+t/2}$ and $J_t=\int_0^tA_s^{-1}\,dB_s$. The <Itô formula> gives $dA_t=-A_t\,dW_t+A_t\,dt$. The independence of the two <Brownian motions> gives $[W,B]=0$, so the product formula yields
$$
\boxed{dX_t=X_t\,dt-X_t\,dW_t-dB_t,\qquad
d[X]_t=(1+X_t^2)\,dt.}
$$
For $f(x)=\arctan x$, $f'(x)=(1+x^2)^{-1}$ and $f''(x)=-2x(1+x^2)^{-2}$. The drift terms cancel in the <Itô formula>:
$$
d\Theta_t=-\frac{X_t}{1+X_t^2}\,dW_t
-\frac1{1+X_t^2}\,dB_t.
$$
Define the rotated <stochastic integral>
$$
Z_t=-\int_0^t\frac{X_s}{\sqrt{1+X_s^2}}\,dW_s
-\int_0^t\frac1{\sqrt{1+X_s^2}}\,dB_s.
$$
It is a <continuous local martingale> with
$$
\langle Z\rangle_t
=\int_0^t\frac{X_s^2+1}{1+X_s^2}\,ds=t.
$$
The <Lévy characterization of Brownian motion> makes $Z$ a <Brownian motion>. Since $\Theta_t\in(-\pi/2,\pi/2)$ and $\cos\Theta_t=(1+X_t^2)^{-1/2}$,
$$
\boxed{d\Theta_t=\cos\Theta_t\,dZ_t.}
$$
This is the <arctangent transform of a two-noise affine diffusion>.
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