Compute the given expression: $\;$ $\sin \left\{\tan^{-1} \left(\dfrac{8}{15}\right) - \sin^{-1} \left(\dfrac{8}{17}\right)\right\}$
Given expression: $\;$ $\sin \left\{\tan^{-1} \left(\dfrac{8}{15}\right) - \sin^{-1} \left(\dfrac{8}{17}\right)\right\}$
$= \sin \left\{\tan^{-1} \left(\dfrac{8}{15}\right)\right\} \cos \left\{\sin^{-1} \left(\dfrac{8}{17}\right)\right\}$
$\hspace{3cm}$ $- \cos \left\{\tan^{-1} \left(\dfrac{8}{15}\right)\right\} \sin \left\{\sin^{-1} \left(\dfrac{8}{17}\right)\right\}$
$\left[\because \;\; \sin \left(A - B\right) = \sin A \cos B - \cos A \sin B\right]$
$= \dfrac{\dfrac{8}{15}}{\sqrt{1 + \left(\dfrac{8}{15}\right)^2}} \times \sqrt{1 - \left(\dfrac{8}{17}\right)^2} - \dfrac{1}{\sqrt{1 + \left(\dfrac{8}{15}\right)^2}} \times \dfrac{8}{17}$
$\left[\because \;\; \sin \left(\tan^{-1} x\right) = \dfrac{x}{\sqrt{1 + x^2}}, \;\; x > 0 \right.$
$\left. \cos \left(\sin^{-1} x\right) = \sqrt{1 - x^2}, \;\; 0 \leq x \leq 1 \right.$
$\left. \sin \left(\sin^{-1} x\right) = x, \;\; x \in \left[-1, 1\right] \right.$
$\left. \cos \left(\tan^{-1} x\right) = \dfrac{1}{\sqrt{1 + x^2}}, \;\; x >0 \right]$
$= \dfrac{8}{17} \times \dfrac{15}{17} - \dfrac{15}{17} \times \dfrac{8}{17}$
$= 0$