Question

Given that tan A= 5/12 and tan B= -4/3 such that A is an acute angle and B is an obtuse angle find the value of,a) sin (A-B)b) cos (A+B)

Answer 1

By trigonometric identity, solve for sin (A-B)

`\begin{gathered} \sin (A-B)=\sin A\cos B-\sin B\cos A \\ \sin (A-B)=((5)/(13))(-(3)/(5))-((4)/(5))((5)/(13)) \\ \sin (A-B)=-(3)/(13)-(4)/(13) \\ \sin (A-B)=-(7)/(13) \end{gathered}`

Therefore, sin(A-B) = -7/13.

Solve for cos(A+B)

`\begin{gathered} \cos (A+B)=\cos A\cos B-\sin A\sin B \\ \cos (A+B)=((12)/(13))(-(3)/(5))-((5)/(13))((4)/(5)) \\ \cos (A+B)=-(36)/(35)-(4)/(13) \\ \cos (A+B)=-(608)/(455) \end{gathered}`

Therefore, cos(A+B) = -608/455.