Question 1

Let $\alpha = \sin^{-1} \left( \frac{3}{5} \right)$ and $\beta = \cos^{-1} \left( \frac{2}{\sqrt{5}} \right)$. Using principal values, $\sin \alpha = \frac{3}{5} \Rightarrow \tan \alpha = \frac{3}{4}$ and $\cos \beta = \frac{2}{\sqrt{5}} \Rightarrow \tan \beta = \frac{1}{2}$. We need to find $\tan(\alpha - 2\beta) = \frac{\tan \alpha - \tan 2\beta}{1 + \tan \alpha \tan 2\beta}$. First, calculate $\tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{3/4} = \frac{4}{3}$. Substituting these values into the subtraction formula: $\tan(\alpha - 2\beta) = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \left( \frac{3}{4} \right) \left( \frac{4}{3} \right)} = \frac{\frac{9 - 16}{12}}{1 + 1} = \frac{-7/12}{2} = -\frac{7}{24}$ Hence, the correct option is (B).