Step 1: Direct evaluation with the answer choices:
The left side turns out to be an increasing function $F(x)=\dfrac{2x^2}{\sqrt{1+x^2}}$, so testing the options is quick once $F$ is known.
Step 2: Derive F:
Use $\sin(P)-\cos(Q)$ with $P=\cos^{-1}x+\tan^{-1}x$ and $Q=\sin^{-1}x+\tan^{-1}x$. Since $\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$, $\sin P = \cos(\sin^{-1}x-\tan^{-1}x)$. The difference then equals $2\sin(\sin^{-1}x)\sin(\tan^{-1}x) = 2x\cdot\frac{x}{\sqrt{1+x^2}}$.
Step 3: Evaluate:
$F(1/\sqrt2) = \dfrac{1}{\sqrt{3/2}} = 0.816$. $F(1/2) = \dfrac{0.5}{1.118} = 0.447 = \dfrac{1}{\sqrt5}$. The target is $\sin(\cot^{-1}2) = 0.447$.
So $x = \frac12$.
Final Answer:
Option (B).
\[ \boxed{\frac{1}{2} \text{ (B)}} \]