Find the value of tan(A+B)

Question

asked Dec 14, 2023 219k views

1 Answer

Anthony L · Answer 1 · 2023-12-21T02:36:03+0000

Answer:

$\tan (A+B) =-(680)/(111)$

Explanation:

$\tan A=(12)/(5)$

$\cos B=(45)/(53)\implies \sec B=(53)/(45)$

Now, by trigonometric identity:

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

$\implies \tan^2 B =\bigg((53)/(45)\bigg)^2-1$

$\implies \tan^2 B =((53)^2-(45)^2)/((45)^2)$

$\implies \tan^2 B =((28)^2)/((45)^2)$

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

$\implies \tan B =\pm(28)/(45)$

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

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

$\implies \tan B =(28)/(45)$

Next,

$\tan (A+B) =(\tan A +\tan B)/(1-\tan A \tan B)$

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

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

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

$\implies \tan (A+B) =((136)/(45))/((-111)/(225))$

$\implies \tan (A+B) =-(680)/(111)$