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 (s) and (t) are in quadrant IV, (\sin s<0) and (\sin t < 0). Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (s): (\sin s=-\sqrt{1-\cos^{2}s}=-\sqrt{1 - (\frac{8}{17})^{2}}=-\sqrt{1-\frac{64}{289}}=-\sqrt{\frac{289 - 64}{289}}=-\sqrt{\frac{225}{289}}=-\frac{15}{17}) For (t): (\sin t=-\sqrt{1-\cos^{2}t}=-\sqrt{1-(\frac{4}{5})^{2}}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{25 - 16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5})

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

Using the identity (\tan\alpha=\frac{\sin\alpha}{\cos\alpha}), for (s): (\tan s=\frac{\sin s}{\cos s}=\frac{-\frac{15}{17}}{\frac{8}{17}}=-\frac{15}{8}) For (t): (\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{3}{5}}{\frac{4}{5}}=-\frac{3}{4})

Step3: Find (\sin(s + t))

Using the formula (\sin(A + B)=\sin A\cos B+\cos A\sin B) (\sin(s + t)=\sin s\cos t+\cos s\sin t) (=(-\frac{15}{17})\times\frac{4}{5}+\frac{8}{17}\times(-\frac{3}{5})) (=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85})

Step4: Find (\tan(s + t))

Using the formula (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}) (\tan(s + t)=\frac{\tan s+\tan t}{1-\tan s\tan t}=\frac{-\frac{15}{8}-\frac{3}{4}}{1-(-\frac{15}{8})\times(-\frac{3}{4})}) 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}) Then (\tan(s + t)=\frac{-\frac{21}{8}}{-\frac{13}{32}}=\frac{21}{8}\times\frac{32}{13}=\frac{84}{13})

Step5: Determine the quadrant of (s + t)

Since (\sin(s + t)=-\frac{84}{85}<0) and (\tan(s + t)=\frac{84}{13}>0) In quadrant III, (\sin\theta<0) and (\tan\theta>0)

Answer:

(a) (\sin(s + t)=-\frac{84}{85}) (b) (\tan(s + t)=\frac{84}{13}) (c) Quadrant III