# Math Help - [SOLVED] Prove: tan(A+45) = (1+tanA)/(1-tanA)

1. ## [SOLVED] Prove: tan(A+45) = (1+tanA)/(1-tanA)

Prove: tan(A+45) = (1+tanA)/(1-tanA)

Please i need help, i know that you start with the RHS ... but i do not know.

Also try to work with only "sin" and "cos" as we havent learnt any other methods yet.

Thanks so much

2. Originally Posted by Jono Mathematician
Prove: tan(A+45) = (1+tanA)/(1-tanA)

Please i need help, i know that you start with the RHS ... but i do not know.

Also try to work with only "sin" and "cos" as we havent learnt any other methods yet.

Thanks so much
Use the sum of angles formula:
$tan(A + 45) = \frac{tan(A) + tan(45)}{1 - tan(A)~tan(45)}$

$= \frac{1 + tan(A)}{1 - tan(A)}$

-Dan

3. I do not understand that ...

i need it to be :

RHS = (1+tanA)/(1-tanA)

=(1+sinA/cosA)/(1-sinA/cosA) etc.

i am not saying the above is wrong, i just need it like this, coz i have no clue wot u said lol

4. Originally Posted by Jono Mathematician
I do not understand that ...

i need it to be :

RHS = (1+tanA)/(1-tanA)

=(1+sinA/cosA)/(1-sinA/cosA) etc.

i am not saying the above is wrong, i just need it like this, coz i have no clue wot u said lol
The only other way I can think to help you with this is to use the sum of angles formula for sine and cosine:
$sin(A + B) = sin(A)~cos(B) + sin(B)~cos(A)$
and
$cos(A + B) = cos(A)~cos(B) - sin(A)~sin(B)$

Then
$tan(A + 45) = \frac{sin(A + 45)}{cos(A + 45)}$

$= \frac{sin(A)~cos(45) + sin(45)~cos(A)}{cos(A)~cos(45) - sin(A)~sin(45)}$
which can be reduced to the needed expression.

I don't think that there is any way to do this problem without using one of these methods.

-Dan

6. Originally Posted by topsquark
$sin(A + B) = sin(A)~cos(B) + sin(B)~cos(A)$
and
$cos(A + B) = cos(A)~cos(B) - sin(A)~sin(B)$

Then
$tan(A + 45) = \frac{sin(A + 45)}{cos(A + 45)}$

$= \frac{sin(A)~cos(45) + sin(45)~cos(A)}{cos(A)~cos(45) - sin(A)~sin(45)}$
The only thing I can think of to do is continue:
$= \frac{\frac{\sqrt{2}}{2}~sin(A) + \frac{\sqrt{2}}{2}~cos(A)}{\frac{\sqrt{2}}{2}~cos( A) - \frac{\sqrt{2}}{2}~sin(A)}$

$= \frac{sin(A) + cos(A)}{cos(A) - sin(A)}$

Now divide numerator and denominator by cos(A).

-Dan

7. Originally Posted by topsquark
Now divide numerator and denominator by cos(A).
Shouldnt we consider the case cos(A) = 0 separately also? I was wondering whether we could do this case without invoking inverse trig functions. I didnt get any idea

8. please can you do the entire sum, like just add the rest on as i have no clue what is going on,

9. Originally Posted by Isomorphism
Shouldnt we consider the case cos(A) = 0 separately also? I was wondering whether we could do this case without invoking inverse trig functions. I didnt get any idea
Since cos(0) = 1 there is no problem with any undefined expressions. Why do you bring that up as a specific case?

-Dan

10. Originally Posted by Jono Mathematician
please can you do the entire sum, like just add the rest on as i have no clue what is going on,
The only thing left to do is divide the numerator and denominator by cos(A). If you can't do this all at once, then take the numerator and denominator separately and do the division. I presume you know that tan(A) = sin(A) / cos(A), right?

-Dan

11. I have spent the last 4 hours doing 15 or so trig "proving identities" and i really cannot get this one, even after u have said that, i am at the last stage with sina+cosA/cosa-sina...

please can you just show me in numbers, i aint good with reading numbers, as in divide numerator by x etc, i need to see it to understand ...thanks
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12. You don't need to use sin or cos anyway, the tan(A+B) formula constitutes a sufficient proof, and looking at the question, that is probably the one that you are supposed to use.

Start on the left and work to the right. Either use the sin and cosine addition formulae

sin(A+B)=sinAcosB+sinBcosA
cos(A+B)=cosAcosB-sinAsinB

and the identity tanx=sinx/cosx

which is the same as saying tan(A+B)=sin(A+B)/cos(A+B)

then the rest should be quite easy, just replace the sin and cos in the fraction with the identities above.

Alternatively and much easier, there is an additional identity,

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

which you can take to be true (it's proved easily by expressing tan as sin/cos but you don't need to do this).