Proving trig equations using addition formulae

[Bonus vir quid]

SOLVED! :)

Homework Statement

The question is as follows;

Prove that (tan(A+B)-tanA)/1+tan(A+B)tanA = tanB

Homework Equations

I'm certain the addition fomulae need to be applied, although I'm not entirely sure how. I genuinely have tried many times!

The Attempt at a Solution

Okay, my first attempt was simply to work with the LHS, rewriting tan(A+B) as (tanA+tanB)/1-tanAtanB on both the top and bottom. I then tried to write tanA as sinA/cosA but this just made the LHS really messy and I wasn't able to cancel anything down.

I then tried to multiply both sides by (1+tan(A+B)tanA), leaving me with (tan(A+B)-tanA) = tanB(1+tan(A+B)tanA). I then expanded the brackets, rewrote all the tans as sin/cos and then attempted to cancel, but this again proved useless as I just ended up with a really really really messy equation.

I've tried this question many times and I still haven't managed to prove it. I am sure this is something very small that I'm missing/forgetting as this is the first trig question that I'm genuinely struggling with. I would appreciate if someone could nudge me in the right direction. Thank you in advance :).

 

[BruceW]

as you say, the key is the addition formula. So try getting tan(A+B) on one side of the equation, and see if the other side agrees with the addition formula.

 

[Bonus vir quid]

as you say, the key is the addition formula. So try getting tan(A+B) on one side of the equation, and see if the other side agrees with the addition formula.

Okay, so I had a go at getting tan(A+B) on one side but I only ended up going in circles again. Perhaps it would help if I showed you how I got lost.

First I moved 1+tan(A+B)tanA over to get
tan(A+B) - tanA = tanB + tan(A+B)tanAtanB ... This is where I suspect I may be going wrong

Then I rearragned to get
tan(A+B) - tan(A+B)tanAtanB = tanA + tanB

Okay, so this is the part where I start losing it and going in circles

Factoring out tan(A+B) I get
tan(A+B)(1-tanAtanB) = tan(A+B)

∴ tanAtanB = 2

And I'm certain that's not correct. Perhaps this makes it easier for you to spot wher I've gone wrong. Thanks again


EDIT: I've managed to crack it, I just forgot to implement the formula - silly me! Thank you so much for pointing me in the right direction though :).

 

[eumyang]

First, you should have clarified in the beginning the original problem, because what you wrote:
(tan(A+B)-tanA)/1+tan(A+B)tanA = tanB
looks like this:
$\frac{\tan (A+B) - \tan A}{1} + \tan (A+B) \tan A = \tan B$

Then I rearragned to get
tan(A+B) - tan(A+B)tanAtanB = tanA + tanB

Okay, so this is the part where I start losing it and going in circles

Factoring out tan(A+B) I get
tan(A+B)(1-tanAtanB) = tan(A+B)

It looks like you changed the right side from "tan A + tan B" to "tan (A+B)". Why? You can't do that. Leave it as it was; you're last step should be
$\tan (A+B)(1 - \tan A \tan B) = \tan A + \tan B$
So what happens next?

EDIT: OP edited his/her previous post while I was writing mine. :)

 

[Bonus vir quid]

First, you should have clarified in the beginning the original problem, because what you wrote:
(tan(A+B)-tanA)/1+tan(A+B)tanA = tanB
looks like this:
$\frac{\tan (A+B) - \tan A}{1} + \tan (A+B) \tan A = \tan B$


It looks like you changed the right side from "tan A + tan B" to "tan (A+B)". Why? You can't do that. Leave it as it was; you're last step should be
$\tan (A+B)(1 - \tan A \tan B) = \tan A + \tan B$
So what happens next?

Apologies for not making the question clear, I forgot to put some bracket in

And even though you were giving instruction for solving a different problem to the one I thought I asked, you managed to help me!

I moved across the (1+tan(A+B)tanA) and then brought the tan(A+B) to one side like you said

I ended up with
tan(A+B)-tan(A+B)tanAtanB = tanB+tanA

Factoring out tan(A+B) I got
tan(A+B)(1-tanAtanB) = tanA + tanB

Rewriting tan(A+B) as tanA + tanB/(1-tanAtanB) I got
tanA + tanB/(1-tanAtanB)*(1-tanAtanB) =tanA+tanB

Cancelling the (1-tanAtanB)'s left me with
tanA+tanB = tanA+tanB

I think that is correct, as I've proved both sides to be equal.

Again, thank you very much.