Find out
[tex]tan(\alpha+\beta)[/tex]Remember that
[tex]tan(\alpha+\beta)=\frac{tan\alpha\text{+}tan\beta}{1-tan\alpha *tan\beta}[/tex]step 1
Find out the cosine of angle alpha
the angle alpha lies on the II quadrant
the value of the cosine of angle alpha is negative
Remember that
[tex]\begin{gathered} sin^2\alpha+cos^2\alpha=1 \\ sin\alpha=\frac{3}{5} \end{gathered}[/tex]substitute
[tex]\begin{gathered} (\frac{3}{5})^2+cos^2\alpha=1 \\ cos^2\alpha=1-\frac{9}{25} \\ \\ cos^2\alpha=\frac{16}{25} \\ cos\alpha=-\frac{4}{5} \end{gathered}[/tex]Find out the tangent of angle alpha
[tex]tan\alpha=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}[/tex]step 2
Find out the value of the sine of the angle beta
The angle beta lies on the III quadrant
so
The value of the sine is negative
Remember that
[tex]\begin{gathered} sin^2\beta+cos^2\beta=1 \\ cos\beta=-\frac{12}{13} \end{gathered}[/tex]substitute
[tex]\begin{gathered} s\imaginaryI n^2\beta+(-\frac{12}{13})^2=1 \\ \\ s\imaginaryI n^2\beta=1-\frac{144}{169} \\ \\ s\imaginaryI n^2\beta=\frac{25}{169} \\ \\ s\imaginaryI n\beta=-\frac{5}{13} \end{gathered}[/tex]Find out the value of the tangent of the angle beta
[tex]tan\beta=\frac{-\frac{5}{13}}{-\frac{12}{13}}=\frac{5}{12}[/tex]step 3
Find out
[tex]tan(\alpha+\beta)=\frac{tan\alpha+\text{t}an\beta}{1-tan\alpha tan\beta}[/tex]substitute given values
[tex]tan(\alpha+\beta)=\frac{(-\frac{3}{4})+(\frac{5}{12})}{1-(-\frac{3}{4})(\frac{5}{12})}[/tex][tex]tan(\alpha+\beta)=\frac{(-3\/4)+(\frac{5}{12})}{1+15\/48)}=\frac{\frac{-9+5}{12}}{\frac{48+15}{48}}=\frac{\frac{-4}{12}}{\frac{33}{48}}=-\frac{4}{12}\div\frac{33}{48}=-\frac{48*4}{33*12}=-\frac{48}{99}[/tex]simplify
