$\text{Considering only the principal values of the inverse trigonometric functions, the value of}$

$\tan \left({\sin }^{-1}\right(\frac{3}{5})-2{\cos }^{-1}(\frac{2}{\sqrt{5}}\left)\right)\text{is}$

The correct option is -7/24.

We need to evaluate the following expression using the principal values of inverse trigonometric functions:

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

Step 1: Define substitution variables for the inverse trigonometric terms.
Let $A={\sin }^{-1}\left(\frac{3}{5}\right)$ and $B={\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)$.

From the principal value branches, both $A$ and $B$ lie in the first quadrant, i.e., $A,B\in (0,\frac{\pi }{2})$.

Step 2: Find $\tan \left(A\right)$ and $\tan \left(B\right)$.

Since $\sin \left(A\right)=\frac{3}{5}$, we have:

$\cos \left(A\right)=\sqrt{1-{\sin }^2\left(A\right)}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$

Thus, $\tan \left(A\right)=\frac{\sin \left(A\right)}{\cos \left(A\right)}=\frac{3/5}{4/5}=\frac{3}{4}$.

Similarly, for $\cos \left(B\right)=\frac{2}{\sqrt{5}}$:

$\sin \left(B\right)=\sqrt{1-{\cos }^2\left(B\right)}=\sqrt{1-\frac{4}{5}}=\frac{1}{\sqrt{5}}$

Thus, $\tan \left(B\right)=\frac{\sin \left(B\right)}{\cos \left(B\right)}=\frac{1/\sqrt{5}}{2/\sqrt{5}}=\frac{1}{2}$.

Step 3: Calculate $\tan \left(2B\right)$ using the double-angle formula for tangent.

$\tan \left(2B\right)=\frac{2\tan \left(B\right)}{1-{\tan }^2\left(B\right)}$

Substituting $\tan \left(B\right)=\frac{1}{2}$:

$\tan \left(2B\right)=\frac{2\left(\frac{1}{2}\right)}{1-{\left(\frac{1}{2}\right)}^2}=\frac{1}{1-\frac{1}{4}}=\frac{1}{3/4}=\frac{4}{3}$

Step 4: Compute $\tan (A-2B)$ using the tangent subtraction formula.

$\tan (A-2B)=\frac{\tan \left(A\right)-\tan \left(2B\right)}{1+\tan \left(A\right)\tan \left(2B\right)}$

Substitute $\tan \left(A\right)=\frac{3}{4}$ and $\tan \left(2B\right)=\frac{4}{3}$:

$\tan (A-2B)=\frac{\frac{3}{4}-\frac{4}{3}}{1+\left(\frac{3}{4}\right)\left(\frac{4}{3}\right)}$

Simplifying the numerator and denominator separately:

$\text{Numerator:}\frac{3}{4}-\frac{4}{3}=\frac{9-16}{12}=-\frac{7}{12}$

$\text{Denominator:}1+1=2$

Therefore:

$\tan (A-2B)=\frac{-7/12}{2}=-\frac{7}{24}$

Thus, the final evaluated value is -7/24.