points details my notes larpcalclimaga8 5.4.047. 0/6 submissions used the exact value of the trigonometric…

Explanation:

Step1: Find cos(u)

Since $\sin^{2}u+\cos^{2}u = 1$, then $\cos^{2}u=1 - \sin^{2}u$. Given $\sin(u)=-\frac{12}{13}$, so $\cos^{2}u=1-\left(-\frac{12}{13}\right)^{2}=1-\frac{144}{169}=\frac{169 - 144}{169}=\frac{25}{169}$. In Quadrant III, $\cos(u)<0$, so $\cos(u)=-\frac{5}{13}$.

Step2: Find tan(u)

We know that $\tan(u)=\frac{\sin(u)}{\cos(u)}$. Substituting $\sin(u)=-\frac{12}{13}$ and $\cos(u)=-\frac{5}{13}$, we get $\tan(u)=\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$.

Step3: Find sin(v)

Since $\sin^{2}v+\cos^{2}v = 1$, then $\sin^{2}v=1 - \cos^{2}v$. Given $\cos(v)=-\frac{15}{17}$, so $\sin^{2}v=1-\left(-\frac{15}{17}\right)^{2}=1-\frac{225}{289}=\frac{289 - 225}{289}=\frac{64}{289}$. In Quadrant III, $\sin(v)<0$, so $\sin(v)=-\frac{8}{17}$.

Step4: Find tan(v)

We know that $\tan(v)=\frac{\sin(v)}{\cos(v)}$. Substituting $\sin(v)=-\frac{8}{17}$ and $\cos(v)=-\frac{15}{17}$, we get $\tan(v)=\frac{-\frac{8}{17}}{-\frac{15}{17}}=\frac{8}{15}$.

Step5: Use the formula for tan(u + v)

The formula for $\tan(A + B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$. Here $A = u$ and $B = v$, so $\tan(u + v)=\frac{\tan(u)+\tan(v)}{1-\tan(u)\tan(v)}$. Substituting $\tan(u)=\frac{12}{5}$ and $\tan(v)=\frac{8}{15}$, we have $\tan(u + v)=\frac{\frac{12}{5}+\frac{8}{15}}{1-\frac{12}{5}\times\frac{8}{15}}$. First, simplify the numerator: $\frac{12}{5}+\frac{8}{15}=\frac{36 + 8}{15}=\frac{44}{15}$. Then simplify the denominator: $1-\frac{12}{5}\times\frac{8}{15}=1-\frac{96}{75}=\frac{75 - 96}{75}=-\frac{21}{75}=-\frac{7}{25}$. So $\tan(u + v)=\frac{\frac{44}{15}}{-\frac{7}{25}}=\frac{44}{15}\times\left(-\frac{25}{7}\right)=-\frac{220}{21}$.

Answer:

$-\frac{220}{21}$