= Solution
Interpret the source's missing variable in the limit arrow as $t\to\infty$. We prove <convergence in distribution> by showing that the angle of the <Brownian motion> becomes uniform on the <circle group>.
Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular <Fourier coefficients> satisfy
$$
\mathbb E[Z_t^m]=e^{-m^2t/2}
\longrightarrow
\begin{cases}1,&m=0,\\0,&m\ne0.\end{cases}
$$
These are precisely the <Fourier coefficients> of the <uniform distribution> on the <unit circle>. Thus <expectations> of every <trigonometric polynomial> converge to its uniform angular average. By the <Stone-Weierstrass theorem>, <trigonometric polynomials> approximate continuous functions on the <unit circle> uniformly. Given an approximation error $\varepsilon$, the discrepancy of <expectations> is at most the polynomial discrepancy plus $2\varepsilon$. Taking $t\to\infty$ and then $\varepsilon\to0$ proves <weak convergence of probability measures> to uniform angular measure.
The <function> $z\mapsto(\operatorname{Im}z)^2$ is continuous on the <unit circle>, so the <continuous mapping theorem> and part (a)(ii) imply
$$
\boxed{\Delta(t)\ \xrightarrow[t\to\infty]{\mathrm d}\ A,
\qquad A\sim\operatorname{Beta}\!\left(\frac12,\frac12\right).}
$$
Equivalently, the limiting <arcsine distribution> has <cumulative distribution function> $2\arcsin\sqrt{x}/\pi$ for $0\leq x\leq1$, and <probability density function> $1/(\pi\sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting <mean>.
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