Simplifying

[rachelmaddie]

Hi. Can someone please check this.

(tan9x-tan5x)/(1+tan9xtan5x)
tan(A -B) = (tanA – tanB)/(1 + tanAtanB)
tan(9x -5x) = (tan9x-tan5x)/(1+tan9xtan5x)
= tan(4x)

 

[Steven G]

Looks good to me. Good job!

 

[lookagain]

Hi. Can someone please check this.
View attachment 20902
(tan9x-tan5x)/(1+tan9xtan5x)
tan(A -B) = (tanA – tanB)/(1 + tanAtanB)
tan(9x -5x) = (tan9x-tan5x)/(1+tan9xtan5x)
= tan(4x)

The problem given to you should have looked like this because those are functions, and the angles are not just lone variables (think the Order of Operations).

\(\displaystyle \dfrac{tan(9x) - tan(5x)}{1 + tan(9x)tan(5x)}\)

Notice how you wrote 4x inside parentheses in your last line.

In keeping with that, your work could look like this:

[tan(9x) - tan(5x)]/[1 + tan(9x)tan(5x)]

tan(A - B) = [tan(A) - tan(B)]/[1 + tan(A)tan(B)]

tan(9x - 5x) = [tan(9x) - tan(5x)]/[1 + tan(9x)tan(5x)]

= tan(4x)