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{3}{5}) (in quadrant II, (\sin s>0)): (\sin s=\sqrt{1-\cos^{2}s}=\sqrt{1 - (-\frac{3}{5})^{2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}) For (t) with (\cos t =-\frac{8}{17}) (in quadrant II, (\sin t>0)): (\sin t=\sqrt{1-\cos^{2}t}=\sqrt{1-(-\frac{8}{17})^{2}}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17})

Step2: Calculate (\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) Substitute (\sin s=\frac{4}{5},\cos s=-\frac{3}{5},\sin t=\frac{15}{17},\cos t=-\frac{8}{17}) (\sin(s + t)=\frac{4}{5}\times(-\frac{8}{17})+(-\frac{3}{5})\times\frac{15}{17}) (=\frac{-32 - 45}{85}=\frac{-77}{85})

Step3: Calculate (\tan s) and (\tan t)

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

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

Using the formula (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}) (\tan(s + t)=\frac{-\frac{4}{3}+(-\frac{15}{8})}{1-(-\frac{4}{3})\times(-\frac{15}{8})}) First, simplify the numerator: (-\frac{4}{3}-\frac{15}{8}=-\frac{32 + 45}{24}=-\frac{77}{24}) Simplify the denominator: (1-\frac{60}{24}=1-\frac{5}{2}=-\frac{3}{2}) Then (\tan(s + t)=\frac{-\frac{77}{24}}{-\frac{3}{2}}=\frac{77}{36})

Step5: Determine the quadrant of (s + t)

Since (\sin(s + t)<0) and (\tan(s + t)>0) In quadrant III, (\sin\theta<0) and (\tan\theta>0)

Answer:

(a) (\sin(s + t)=-\frac{77}{85}) (b) (\tan(s + t)=\frac{77}{36}) (c) Quadrant III