Answer:
Step-by-step explanation:
[tex]sin ~a=\frac{12}{13} ,\frac{\pi}{2} <a<\pi\\a ~is~in~2nd~quadrant.\\sin~b=\frac{3}{5} ,0<b<\frac{\pi}{2} \\b~is~in~1st~quadrant.\\cos~b=\sqrt{1-sin ^2b} \\=\sqrt{1-(\frac{3}{5} )^2} \\=\sqrt{\frac{25-9}{25} } \\=\sqrt{\frac{16}{25} } \\=\frac{4}{5} ,\\cos~a=-\sqrt{1-sin^2 a} \\=-\sqrt{1-(\frac{12}{13} )^2} \\=-\sqrt{\frac{169-144}{169} } \\=-\sqrt{\frac{25}{169} } \\=-\frac{5}{13} \\cos(a-b)=cos~acos~b+sin~a sin~b\\=\frac{-5}{13}* \frac{4}{5} +\frac{12}{13}* \frac{3}{5} \\=?[/tex]
