Question

Find the value of tan(A+B)

Answer 1

`\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)`