use the given information to find (a) \\( \\sin (s+t) \\), (b) \\( \\tan (s+t) \\), and (c) the quadrant of…

Explanation:

Step1: Find (\sin s) and (\sin t)

Since (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (s) with (\cos s=\frac{8}{17}) and (s) in quadrant IV ((\sin s<0)), we have (\sin s=-\sqrt{1 - \cos^{2}s}=-\sqrt{1-\left(\frac{8}{17}\right)^{2}}=-\sqrt{\frac{289 - 64}{289}}=-\frac{15}{17}). For (t) with (\cos t=\frac{4}{5}) and (t) in quadrant IV ((\sin t<0)), we have (\sin t=-\sqrt{1-\cos^{2}t}=-\sqrt{1 - \left(\frac{4}{5}\right)^{2}}=-\sqrt{\frac{25-16}{25}}=-\frac{3}{5}).

Step2: Find (\tan s) and (\tan t)

(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}), so (\tan s=\frac{\sin s}{\cos s}=\frac{-\frac{15}{17}}{\frac{8}{17}}=-\frac{15}{8}), (\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{3}{5}}{\frac{4}{5}}=-\frac{3}{4}).

Step3: Use the formula (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B})

Substitute (A = s) and (B = t) into the formula (\tan(s + t)=\frac{\tan s+\tan t}{1-\tan s\tan t}). (\tan s=-\frac{15}{8}), (\tan t=-\frac{3}{4}), then (\tan(s + t)=\frac{-\frac{15}{8}-\frac{3}{4}}{1-\left(-\frac{15}{8}\right)\left(-\frac{3}{4}\right)}). First, simplify the numerator: (-\frac{15}{8}-\frac{3}{4}=-\frac{15 + 6}{8}=-\frac{21}{8}). Second, simplify the denominator: (1-\frac{45}{32}=\frac{32-45}{32}=-\frac{13}{32}). So (\tan(s + t)=\frac{-\frac{21}{8}}{-\frac{13}{32}}=\frac{21\times4}{13}=\frac{84}{13}).

Answer:

(\frac{84}{13})