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)

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

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

Solve for cos(A+B)

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

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