1. **Problem statement:** (a) Find $\tan\left(\arcsin\left(\frac{8}{x}\right)\right)$. (b) Find $\sec\left(\arctan\left(\frac{x}{7}\right)\right)$. --- 2. **Recall the definitions and formulas:** - For an angle $\theta$, if $\sin \theta = y$, then $\tan \theta = \frac{\sin \theta}{\cos \theta}$. - Also, $\cos \theta = \sqrt{1 - \sin^2 \theta}$ when $\theta$ is in the principal range. - For $\arctan z = \phi$, $\tan \phi = z$, and $\sec \phi = \frac{1}{\cos \phi} = \sqrt{1 + \tan^2 \phi}$. --- 3. **Part (a):** Let $\theta = \arcsin\left(\frac{8}{x}\right)$, so $\sin \theta = \frac{8}{x}$. Calculate $\cos \theta$: $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{8}{x}\right)^2} = \sqrt{1 - \frac{64}{x^2}} = \sqrt{\frac{x^2 - 64}{x^2}} = \frac{\sqrt{x^2 - 64}}{|x|}.$$ Then, $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{8}{x}}{\frac{\sqrt{x^2 - 64}}{|x|}} = \frac{8}{x} \cdot \frac{|x|}{\sqrt{x^2 - 64}} = \frac{8 |x|}{x \sqrt{x^2 - 64}}.$$ Since $\frac{|x|}{x} = \operatorname{sgn}(x)$, the expression simplifies to $$\tan\left(\arcsin\left(\frac{8}{x}\right)\right) = \frac{8}{\sqrt{x^2 - 64}} \cdot \operatorname{sgn}(x).$$ --- 4. **Part (b):** Let $\phi = \arctan\left(\frac{x}{7}\right)$, so $\tan \phi = \frac{x}{7}$. Recall that $$\sec \phi = \sqrt{1 + \tan^2 \phi} = \sqrt{1 + \left(\frac{x}{7}\right)^2} = \sqrt{1 + \frac{x^2}{49}} = \sqrt{\frac{49 + x^2}{49}} = \frac{\sqrt{x^2 + 49}}{7}.$$ Therefore, $$\sec\left(\arctan\left(\frac{x}{7}\right)\right) = \frac{\sqrt{x^2 + 49}}{7}.$$ --- **Final answers:** (a) $\tan\left(\arcsin\left(\frac{8}{x}\right)\right) = \frac{8}{\sqrt{x^2 - 64}} \cdot \operatorname{sgn}(x)$. (b) $\sec\left(\arctan\left(\frac{x}{7}\right)\right) = \frac{\sqrt{x^2 + 49}}{7}$.