Let us draw a sketch to understand the question
Since A is an acute angle, then it lies on the 1st quadrant
Since B is a reflex angle, then it lies on the 3rd quadrant
In the first quadrant, all ratios are +ve
In the 3rd quadrant sin and cos - ve but the tan is +ve
We will use the identity
[tex]\sin ^2B+\cos ^2B=1[/tex]To find the value of sin B because tan B = sin B/cos B
Since cos B = -4/5, then
[tex]\begin{gathered} \sin ^2B+(-\frac{4}{5})^2=1 \\ \sin ^2B+\frac{16}{25}=1 \end{gathered}[/tex]Subtract 16/25 from both sides
[tex]\begin{gathered} \sin ^2B+\frac{16}{25}-\frac{16}{25}=1-\frac{16}{25} \\ \sin ^2B=\frac{9}{25} \end{gathered}[/tex]Take a square root for both sides
[tex]\begin{gathered} \sqrt[]{\sin^2B}=\pm\sqrt[]{\frac{9}{25}} \\ \sin B=-\frac{3}{5} \end{gathered}[/tex]We take the negative value because sin in the 3rd quadrant is -ve
Divide sin B by cos B to find tan B
[tex]\begin{gathered} \tan B=\frac{\sin B}{\cos B} \\ \tan B=\frac{-\frac{3}{5}}{-\frac{4}{5}} \\ \tan B=\frac{3}{4} \end{gathered}[/tex]a) tan B = 3/4
Since sin(A + b) = sinA cosB + SinB cosA --------(1)
Then we have to find the value of cos A, we will use the identity above
[tex]\begin{gathered} \sin ^2A+\cos ^2A=1 \\ (\frac{12}{13})^2+\cos ^2A=1 \\ \frac{144}{169}+\cos ^2A=1 \end{gathered}[/tex]Subtract 144/169 from both sides to find the value of cos^2A
[tex]\begin{gathered} \frac{144}{169}-\frac{144}{169}+\cos ^2A=1-\frac{144}{169} \\ \cos ^2A=\frac{25}{169} \end{gathered}[/tex]Take a square root for both sides, then
[tex]\begin{gathered} \sqrt[]{\cos^2A}=\pm\sqrt[]{\frac{25}{169}} \\ \cos A=\frac{5}{13} \end{gathered}[/tex]Substitute the values of sin A, cos A, and sin B, cos B in expression (1) above to find sin (A + B)
[tex]\begin{gathered} \sin (A+B)=\sin A\cos B+\sin B\cos A \\ \sin (A+B)=\frac{12}{13}\times(-\frac{4}{5})+(-\frac{3}{5})\times(\frac{5}{13}) \end{gathered}[/tex]Simplify it
[tex]\begin{gathered} \sin (A+B)=\frac{-48}{65}+(\frac{-15}{65}) \\ \sin (A+B)=-\frac{63}{65} \end{gathered}[/tex]b) sin(A + b) = -63/65
