Considering only the principal values of the inverse trigonometric functions, the value of $$\tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$$ is

Let $$\alpha = \sin^{-1}\!\left(\frac{3}{5}\right)$$ and $$\beta = \cos^{-1}\!\left(\frac{2}{\sqrt{5}}\right)$$. Both are principal values, so $$-\frac{\pi}{2}\le \alpha\le \frac{\pi}{2}$$ and $$0\le \beta\le \pi$$.

Step 1: Evaluate $$\sin\alpha, \cos\alpha, \tan\alpha$$
Given $$\sin\alpha = \frac{3}{5}$$, hence $$\cos\alpha = \sqrt{1-\sin^2\alpha}= \sqrt{1-\left(\frac{3}{5}\right)^2}= \sqrt{\frac{16}{25}}=\frac{4}{5}$$ (positive in the principal range).
Therefore $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}= \frac{\tfrac{3}{5}}{\tfrac{4}{5}} = \frac{3}{4}$$.

Step 2: Evaluate $$\sin\beta, \cos\beta, \tan\beta$$
Given $$\cos\beta = \frac{2}{\sqrt{5}}$$, so $$\sin\beta = \sqrt{1-\cos^2\beta}= \sqrt{1-\frac{4}{5}}= \sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}}$$ (positive because $$0\lt\beta\lt\frac{\pi}{2}$$).
Hence $$\tan\beta = \frac{\sin\beta}{\cos\beta}= \frac{\tfrac{1}{\sqrt{5}}}{\tfrac{2}{\sqrt{5}}}= \frac{1}{2}$$.

Step 3: Compute $$\tan(2\beta)$$
Use the double-angle formula $$\tan(2\beta)=\frac{2\tan\beta}{1-\tan^{2}\beta}$$.
Substituting $$\tan\beta=\frac{1}{2}$$ gives $$\tan(2\beta)=\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^2}= \frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$$.

Step 4: Compute $$\tan\!\bigl(\alpha-2\beta\bigr)$$
The subtraction formula is $$\tan\!\bigl(\alpha-2\beta\bigr)=\frac{\tan\alpha-\tan(2\beta)}{1+\tan\alpha\,\tan(2\beta)}.$$

Numerator: $$\tan\alpha-\tan(2\beta)=\frac{3}{4}-\frac{4}{3}= \frac{9-16}{12}= -\frac{7}{12}.$$ Denominator: $$1+\tan\alpha\,\tan(2\beta)=1+\frac{3}{4}\cdot\frac{4}{3}=1+1=2.$$

Thus $$\tan\!\bigl(\alpha-2\beta\bigr)=\frac{-\tfrac{7}{12}}{2}= -\frac{7}{24}.$$

Therefore the required value is $$\frac{-7}{24}$$.

Option B which is: $$\frac{-7}{24}$$