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 (\cos t)

Since (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (s) with (\cos s=-\frac{12}{13}) (in quadrant III), (\sin s=-\sqrt{1-\cos^{2}s}=-\sqrt{1 - (\frac{- 12}{13})^{2}}=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{169 - 144}{169}}=-\frac{5}{13}) For (t) with (\sin t=-\frac{4}{5}) (in quadrant III), (\cos t=-\sqrt{1-\sin^{2}t}=-\sqrt{1-(\frac{-4}{5})^{2}}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{25 - 16}{25}}=-\frac{3}{5})

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{5}{13},\cos t=-\frac{3}{5},\cos s=-\frac{12}{13},\sin t =-\frac{4}{5}) (\sin(s + t)=(-\frac{5}{13})(-\frac{3}{5})+(-\frac{12}{13})(-\frac{4}{5})=\frac{15}{65}+\frac{48}{65}=\frac{15 + 48}{65}=\frac{63}{65})

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

(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}), so (\tan s=\frac{\sin s}{\cos s}=\frac{-\frac{5}{13}}{-\frac{12}{13}}=\frac{5}{12}), (\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{4}{5}}{-\frac{3}{5}}=\frac{4}{3})

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{\tan s+\tan t}{1-\tan s\tan t}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times\frac{4}{3}}=\frac{\frac{5 + 16}{12}}{1-\frac{20}{36}}=\frac{\frac{21}{12}}{\frac{36- 20}{36}}=\frac{\frac{21}{12}}{\frac{16}{36}}=\frac{21}{12}\times\frac{36}{16}=\frac{63}{16})

Step5: Determine the quadrant of (s + t)

Since (\sin(s + t)=\frac{63}{65}>0) and (\tan(s + t)=\frac{63}{16}>0) In quadrant I, (\sin\theta>0) and (\tan\theta>0)

Answer:

(a) (\frac{63}{65}) (b) (\frac{63}{16}) (c) Quadrant I