Expert: Sherry Wallin - 5/3/2010

Question
hello Ahmed, i am stuck on an identity question, which is as follows...
Use the identity tan(A-B) = (tanA - tanB)/(1+tanAtanB) with A= (r+1)x and B= rx to ahow that

tanrxtan(r+1)x= (tan(r+1)x/tanx)-(tanrx/tanx)-1
thankyou in advance.

Answer
Jason~
This is not Ahmed, it is Math Prof. Now let's look at the identity
tan(A-B) = (tanA - tanB)/(1+tanAtanB) and massage it some:
multipy both sides by (1+tanAtanB) getting:

(1+tanAtanB)tan(A-B) =  (1+tanAtanB)(tanA - tanB)/(1+tanAtanB)->

(1+tanAtanB)tan(A-B)= (tanA - tanB)  Now divide both sides by tan(A-B) ->

(1+tanAtanB)tan(A-B)/tan(A-B) = (tanA - tanB)/tan(A-B)->

(1+tanAtanB) = (tanA - tanB)/tan(A-B) Now subtract 1 from both sides ->
1+tanAtanB - 1 = (tanA - tanB)/tan(A-B) - 1 ->
tanAtanB = [(tanA - tanB)/tan(A-B)] - 1 -> you can multiply in any order so this is equivalent to:
tanBtanA = [(tanA - tanB)/tan(A-B)] - 1  Now substitute what you have above for A and B ->
tanrxtan(r+1)x = [(tan(r+1)x - tanrx)/[tan((r+1)x-rx)]]-1 ->
tanrxtan(r+1)x = [(tan(r+1)x - tanrx)/[tan(rx+x-rx)]]-1 ->
tanrxtan(r+1)x = [(tan(r+1)x - tanrx)/(tanx)]-1 ->
tanrxtan(r+1)x = [(tan(r+1)x]/(tanx)  - (tanrx/tanx)-1 ->
tanrxtan(r+1)x= (tan(r+1)x/tanx)-(tanrx/tanx)-1  as desired!

Math Prof