For $x > 0$, if $\sin(\cos^{-1}x + \tan^{-1}x) - \cos(\sin^{-1}x + \tan^{-1}x) = \sin(\cot^{-1}2)$ then $x = $

Given equation: $\sin(\cos^{-1}x + \tan^{-1}x) - \cos(\sin^{-1}x + \tan^{-1}x) = \sin(\cot^{-1}2)$

Let $\alpha = \cos^{-1}x$ and $\gamma = \tan^{-1}x$.

We know that $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$, so $\sin^{-1}x = \dfrac{\pi}{2} - \alpha$.

The second term on the left hand side becomes $\cos\left(\dfrac{\pi}{2} - \alpha + \gamma\right) = \sin(\alpha - \gamma)$.

The left hand side simplifies to $\sin(\alpha + \gamma) - \sin(\alpha - \gamma) = 2 \cos\alpha \sin\gamma$.

Since $\alpha = \cos^{-1}x$, we have $\cos\alpha = x$.

Since $\gamma = \tan^{-1}x$, we have $\sin\gamma = \dfrac{x}{\sqrt{1+x^2}}$.

Thus, the left hand side is $2x \left( \dfrac{x}{\sqrt{1+x^2}} \right) = \dfrac{2x^2}{\sqrt{1+x^2}}$.

For the right hand side, let $\theta = \cot^{-1}2$, then $\cot\theta = 2$, which gives $\sin\theta = \dfrac{1}{\sqrt{1+\cot^2\theta}} = \dfrac{1}{\sqrt{5}}$.

Equating both sides:

$\dfrac{2x^2}{\sqrt{1+x^2}} = \dfrac{1}{\sqrt{5}}$

Squaring both sides:

$\dfrac{4x^4}{1+x^2} = \dfrac{1}{5}$

$20x^4 = 1 + x^2$

$20x^4 - x^2 - 1 = 0$

$(5x^2 + 1)(4x^2 - 1) = 0$

Since $x^2 > 0$, we must have $4x^2 - 1 = 0 \Rightarrow x^2 = \dfrac{1}{4}$.

Given $x > 0$, we get $x = \dfrac{1}{2}$.

Answer: $\dfrac{1}{2}$