Question

Find the value of tan(A+B)

Answer 1

Answer:

$\tan (A+B) =-\frac{680}{111}$

Step-by-step explanation:

$\tan A=\frac{12}{5}$

$\cos B=\frac{45}{53}\implies \sec B=\frac{53}{45}$

Now, by trigonometric identity:

$\tan^2 B =\sec^2B-1$

$\implies \tan^2 B =\bigg(\frac{53}{45}\bigg)^2-1$

$\implies \tan^2 B =\frac{(53)^2-(45)^2}{(45)^2}$

$\implies \tan^2 B =\frac{(28)^2}{(45)^2}$

$\implies \tan B =\pm\sqrt{\frac{(28)^2}{(45)^2}}$

$\implies \tan B =\pm\frac{28}{45}$

It is given that: Angles A and B are in quadrant I.

---------------------> Angles A and B would be positive.

$\implies \tan B =\frac{28}{45}$

Next,

$\tan (A+B) =\frac{\tan A +\tan B}{1-\tan A \tan B}$

$\implies \tan (A+B) =\frac{\frac{12}{5} +\frac{28}{45}}{1-\bigg(\frac{12}{5}\bigg) \bigg(\frac{28}{45}\bigg)}$

$\implies \tan (A+B) =\frac{\frac{108}{45} +\frac{28}{45}}{1-\bigg(\frac{336}{225}\bigg)}$

$\implies \tan (A+B) =\frac{\frac{136}{45}}{\bigg(\frac{225-336}{225}\bigg)}$

$\implies \tan (A+B) =\frac{\frac{136}{45}}{\frac{-111}{225}}$

$\implies \tan (A+B) =-\frac{680}{111}$