Considering only the principal values of the inverse trigonometric...

Let $E = \tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$.

Evaluate the inverse trigonometric expressions: Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$, so $\sin\alpha = \frac{3}{5}$ and $\tan\alpha = \frac{3}{4}$. Let $\beta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)$, so $\cos\beta = \frac{2}{\sqrt{5}}$ and $\tan\beta = \frac{1}{2}$.

Compute the double angle: Using $\tan(2\beta) = \frac{2\tan\beta}{1 - \tan^2\beta}$, we obtain $\tan(2\beta) = \frac{2\cdot\frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{4}{3}$.

Apply the tangent subtraction formula: $E = \tan(\alpha - 2\beta) = \frac{\tan\alpha - \tan(2\beta)}{1 + \tan\alpha\tan(2\beta)} = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \frac{3}{4}\cdot\frac{4}{3}} = \frac{-\frac{7}{12}}{2} = -\frac{7}{24}$.

Final result: $\boxed{-\frac{7}{24}}$