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

Explanation:

Step1: Find $\cos s$ and $\cos t$

Using the identity $\sin^{2}\theta+\cos^{2}\theta = 1$. For $s$ (quadrant II, $\cos s<0$): $\cos s=-\sqrt{1 - \sin^{2}s}=-\sqrt{1-\left(\frac{5}{7}\right)^{2}}=-\sqrt{\frac{49 - 25}{49}}=-\frac{2\sqrt{6}}{7}$ For $t$ (quadrant IV, $\cos t>0$): $\cos t=\sqrt{1 - \sin^{2}t}=\sqrt{1-\left(-\frac{6}{7}\right)^{2}}=\sqrt{\frac{49 - 36}{49}}=\frac{\sqrt{13}}{7}$

Step2: Find $\tan s$ and $\tan t$

Using the formula $\tan\theta=\frac{\sin\theta}{\cos\theta}$. $\tan s=\frac{\sin s}{\cos s}=\frac{\frac{5}{7}}{-\frac{2\sqrt{6}}{7}}=-\frac{5}{2\sqrt{6}}=-\frac{5\sqrt{6}}{12}$ $\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{6}{7}}{\frac{\sqrt{13}}{7}}=-\frac{6}{\sqrt{13}}=-\frac{6\sqrt{13}}{13}$

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

Substitute $A = s$ and $B = t$: $\tan(s + t)=\frac{\tan s+\tan t}{1-\tan s\tan t}$ $=\frac{-\frac{5\sqrt{6}}{12}-\frac{6\sqrt{13}}{13}}{1-\left(-\frac{5\sqrt{6}}{12}\right)\left(-\frac{6\sqrt{13}}{13}\right)}$ First, find a common denominator for the numerator: Numerator: $-\frac{5\sqrt{6}\times13}{12\times13}-\frac{6\sqrt{13}\times12}{13\times12}=-\frac{65\sqrt{6}+72\sqrt{13}}{156}$ Denominator: $1-\frac{5\sqrt{6}\times6\sqrt{13}}{12\times13}=1-\frac{30\sqrt{78}}{156}=\frac{156 - 30\sqrt{78}}{156}$ $\tan(s + t)=\frac{-\frac{65\sqrt{6}+72\sqrt{13}}{156}}{\frac{156 - 30\sqrt{78}}{156}}=\frac{-(65\sqrt{6}+72\sqrt{13})}{156 - 30\sqrt{78}}$ Rationalize the denominator: Multiply numerator and denominator by $156 + 30\sqrt{78}$ After simplification (using $(a - b)(a + b)=a^{2}-b^{2}$ in the denominator and expanding the numerator), we get: $\tan(s + t)=\frac{-(65\sqrt{6}+72\sqrt{13})(156 + 30\sqrt{78})}{156^{2}-(30\sqrt{78})^{2}}$ Another way: We know that $\sin(s + t)=\frac{5\sqrt{13}+12\sqrt{6}}{49}$ and $\cos(s + t)=\cos s\cos t-\sin s\sin t$ $\cos(s + t)=\left(-\frac{2\sqrt{6}}{7}\right)\times\frac{\sqrt{13}}{7}-\frac{5}{7}\times\left(-\frac{6}{7}\right)=\frac{-2\sqrt{78}+30}{49}$ Then $\tan(s + t)=\frac{\sin(s + t)}{\cos(s + t)}=\frac{5\sqrt{13}+12\sqrt{6}}{-2\sqrt{78}+30}=\frac{5\sqrt{13}+12\sqrt{6}}{30 - 2\sqrt{78}}=\frac{(5\sqrt{13}+12\sqrt{6})(30 + 2\sqrt{78})}{(30 - 2\sqrt{78})(30 + 2\sqrt{78})}$ $=\frac{150\sqrt{13}+10\sqrt{13\times78}+360\sqrt{6}+24\sqrt{6\times78}}{900-4\times78}$ $=\frac{150\sqrt{13}+10\sqrt{1014}+360\sqrt{6}+24\sqrt{468}}{900 - 312}$ $=\frac{150\sqrt{13}+10\sqrt{1014}+360\sqrt{6}+24\sqrt{468}}{588}$ Simplify further: $\sqrt{1014}=\sqrt{2\times3\times13^{2}} = 13\sqrt{6}$, $\sqrt{468}=\sqrt{4\times117}=\sqrt{4\times9\times13}=6\sqrt{13}$ $\tan(s + t)=\frac{150\sqrt{13}+10\times13\sqrt{6}+360\sqrt{6}+24\times6\sqrt{13}}{588}$ $=\frac{(150\sqrt{13}+144\sqrt{13})+(130\sqrt{6}+360\sqrt{6})}{588}$ $=\frac{294\sqrt{13}+490\sqrt{6}}{588}=\frac{42\times7\sqrt{13}+70\times7\sqrt{6}}{84\times7}=\frac{42\sqrt{13}+70\sqrt{6}}{84}=\frac{21\sqrt{13}+35\sqrt{6}}{42}=\frac{3\sqrt{13}+5\sqrt{6}}{6}$

Answer:

$\frac{3\sqrt{13}+5\sqrt{6}}{6}$