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

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 I), we have (\sin s=\sqrt{1 - \cos^{2}s}=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{169 - 144}{169}}=\frac{5}{13}). For (t) with (\sin t=\frac{3}{5}) (in quadrant I), we have (\cos t=\sqrt{1-\sin^{2}t}=\sqrt{1 - \left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}).

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

Using the formula (\tan\alpha=\frac{\sin\alpha}{\cos\alpha}), (\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{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+\tan t=\frac{5}{12}+\frac{3}{4}=\frac{5 + 9}{12}=\frac{14}{12}=\frac{7}{6}). (\tan s\tan t=\frac{5}{12}\times\frac{3}{4}=\frac{5}{16}). (1-\tan s\tan t=1-\frac{5}{16}=\frac{16 - 5}{16}=\frac{11}{16}). So (\tan(s + t)=\frac{\frac{7}{6}}{\frac{11}{16}}=\frac{7\times16}{6\times11}=\frac{56}{33}).

Answer:

(\tan(s + t)=\frac{56}{33})