Answer:
[tex]\tan (A+B) =-\frac{680}{111}[/tex]
Step-by-step explanation:
[tex]\tan A=\frac{12}{5}[/tex]
[tex]\cos B=\frac{45}{53}\implies \sec B=\frac{53}{45}[/tex]
Now, by trigonometric identity:
[tex]\tan^2 B =\sec^2B-1[/tex]
[tex]\implies \tan^2 B =\bigg(\frac{53}{45}\bigg)^2-1[/tex]
[tex]\implies \tan^2 B =\frac{(53)^2-(45)^2}{(45)^2}[/tex]
[tex]\implies \tan^2 B =\frac{(28)^2}{(45)^2}[/tex]
[tex]\implies \tan B =\pm\sqrt{\frac{(28)^2}{(45)^2}}[/tex]
[tex]\implies \tan B =\pm\frac{28}{45}[/tex]
It is given that: Angles A and B are in quadrant I.
---------------------> Angles A and B would be positive.
[tex]\implies \tan B =\frac{28}{45}[/tex]
Next,
[tex]\tan (A+B) =\frac{\tan A +\tan B}{1-\tan A \tan B}[/tex]
[tex]\implies \tan (A+B) =\frac{\frac{12}{5} +\frac{28}{45}}{1-\bigg(\frac{12}{5}\bigg) \bigg(\frac{28}{45}\bigg)}[/tex]
[tex]\implies \tan (A+B) =\frac{\frac{108}{45} +\frac{28}{45}}{1-\bigg(\frac{336}{225}\bigg)}[/tex]
[tex]\implies \tan (A+B) =\frac{\frac{136}{45}}{\bigg(\frac{225-336}{225}\bigg)}[/tex]
[tex]\implies \tan (A+B) =\frac{\frac{136}{45}}{\frac{-111}{225}}[/tex]
[tex]\implies \tan (A+B) =-\frac{680}{111}[/tex]
