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

$\tan\left(\sin^{-1}\dfrac{3}{5}-2\cos^{-1}\dfrac{2}{\sqrt{5}}\right)$ $\text{principal values only}$ $\text{To find: the value of the expression}$ $(1)\ \dfrac{7}{24} \quad (2)\ -\dfrac{7}{24} \quad (3)\ -\dfrac{5}{24} \quad (4)\ \dfrac{5}{24}$ $\text{Plan: } A=\sin^{-1}\dfrac35,\quad B=\cos^{-1}\dfrac{2}{\sqrt5}$ $\text{find } \tan A,\ \tan B,\ \tan 2B$ $\text{then } \tan(A-2B)$ $A=\sin^{-1}\dfrac{3}{5}\ \Rightarrow\ \sin A=\dfrac{3}{5},\ A\in\left(0,\dfrac{\pi}{2}\right)$ $\cos A=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}$ $\tan A=\dfrac{\sin A}{\cos A}=\dfrac{3}{4}\ \Rightarrow\ A=\tan^{-1}\dfrac{3}{4}$ $B=\cos^{-1}\dfrac{2}{\sqrt{5}}\ \Rightarrow\ \cos B=\dfrac{2}{\sqrt{5}},\ B\in\left(0,\dfrac{\pi}{2}\right)$ $\sin B=\sqrt{1-\dfrac{4}{5}}=\dfrac{1}{\sqrt{5}}$ $\tan B=\dfrac{1/\sqrt{5}}{2/\sqrt{5}}=\dfrac{1}{2}\ \Rightarrow\ B=\tan^{-1}\dfrac{1}{2}$ $\tan 2B=\dfrac{2\tan B}{1-\tan^{2}B}=\dfrac{2\cdot\frac12}{1-\frac14}=\dfrac{4}{3}$ $A-2B=\tan^{-1}\dfrac{3}{4}-\tan^{-1}\dfrac{4}{3}$ $\tan(A-2B)=\dfrac{\tan A-\tan 2B}{1+\tan A\,\tan 2B}=\dfrac{\frac34-\frac43}{1+\frac34\cdot\frac43}$ $=\dfrac{\frac{9-16}{12}}{1+1}=\dfrac{-7/12}{2}=-\dfrac{7}{24}$ $-\dfrac{7}{24}\ \Rightarrow\ \text{option }(2)$ $\boxed{-\dfrac{7}{24}}\quad\Rightarrow\ (2)$