Considering only the principal values of the inverse trigonometric functions, the value of
$$\tan\left(\sin^{-1}\left(\dfrac{3}{5}\right) - 2\cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)\right)$$
is
Step-by-Step Solution
Key Concept: Using trigonometric identities to convert inverse trigonometric functions to their tangent equivalents, and applying the tangent double-angle and difference formulas.
Let $A = \sin^{-1}\left(\dfrac{3}{5}\right) \implies \tan A = \dfrac{3}{4}$, with $A \in \left(0, \dfrac{\pi}{2}\right)$.
Let $B = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right) \implies \tan B = \dfrac{1}{2}$, with $B \in \left(0, \dfrac{\pi}{2}\right)$.
We want to evaluate $\tan(A - 2B)$:
$$\tan(2B) = \dfrac{2\tan B}{1 - \tan^2 B} = \dfrac{2(1/2)}{1 - 1/4} = \dfrac{1}{3/4} = \dfrac{4}{3}$$
Now, apply the difference formula for tangent:
$$\tan(A - 2B) = \dfrac{\tan A - \tan(2B)}{1 + \tan A \tan(2B)} = \dfrac{\dfrac{3}{4} - \dfrac{4}{3}}{1 + \left(\dfrac{3}{4}\right)\left(\dfrac{4}{3}\right)} = \dfrac{\dfrac{9 - 16}{12}}{1 + 1} = \dfrac{-7/12}{2} = -\dfrac{7}{24}$$
Thus, the correct option is B.
Correct Answer: B