# 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 :).

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Homework Helper
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 :).

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Homework Helper
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.