The given angles are sin A =3/5 and tan B= -12/5 where A is an acute angle and B is an obtuse angle.
a. To determine tan A
First find the cos A with the help of trigonometric property
[tex]\cos A=\sqrt[]{1-\sin ^2A}[/tex][tex]\cos A=\sqrt[]{1-\frac{9}{25}}=\frac{4}{5}[/tex][tex]\tan A=\frac{\sin A}{\cos A}=\frac{3}{4}[/tex]b. To determine cos(A+B)
[tex]\cos (A+B)=\cos A\cos B-\sin A\text{ sin B}[/tex]We have to find sin B and cosB, it is given that tan B= -12/5
[tex]\sec ^2B=1+tan^2B[/tex][tex]\sec ^2B=1+\frac{144}{25}=\frac{169}{25}[/tex][tex]\sec B=\frac{13}{5}[/tex]Then the value of cos B is 5/13.
Determine the value of sinB
[tex]\sin B=\sqrt[]{1-\frac{25}{169}}=\frac{12}{13}[/tex]The value of cos (A+B) i
