Question

Question #7d769

Answer 1

`tan(A+B)-tan(A-B)=(2tanB)/(1-tan^2Atan^2B)`

Explanation:

Use the trigonometric identity
`tan(a+-b)=(tan(a)+-tan(b))/(1+--tan(a)tan(b))`

So,
`tan(A+B)=(tanA+tanB)/(1-tanAtanB)`
`tan(A-B)=(tanA-tanB)/(1+tanAtanB)`
We will substitute:
`a=tanA` and `b=tanB`

So, `tan(A+B)-tan(A-B)=(a+b)/(1-ab)-(a-b)/(1+ab)`

`(a+b)/(1-ab)-(a-b)/(1+ab)=((a+b)(1+ab)-(a-b)(1-ab))/(1-a^2b^2)`
`=(2b)/(1-a^2b^2)`

Sub back:
`tan(A+B)-tan(A-B)=(2tanB)/(1-tan^2Atan^2B)`

Well... that's ugly but whatever.