Question #7d769

1 Answer
May 24, 2017

#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.