# Prove tan

• Feb 12th 2010, 09:24 AM
gary223
Prove tan
Prove that $\displaystyle tan(\theta+45degrees) = \frac{1+tan\theta}{1-tan\theta}$
• Feb 12th 2010, 09:29 AM
Krizalid
straightforward, just use the formula for $\displaystyle \tan(x+y).$
• Feb 12th 2010, 09:31 AM
gary223
I've never done this type question before, could you post how its done please. I know I have to use the formula but how is it done?
• Feb 12th 2010, 09:32 AM
Krizalid
$\displaystyle \tan(x+y)=\frac{\tan x+\tan y}{1-\tan(x)\tan(y)}.$
• Feb 12th 2010, 09:38 AM
gary223
So what is x and y? Is it $\displaystyle 1+tan\theta$ and $\displaystyle 1-tan\theta$?
• Feb 12th 2010, 09:43 AM
skeeter
Quote:

Originally Posted by gary223
So what is x and y? Is it $\displaystyle 1+tan\theta$ and $\displaystyle 1-tan\theta$?

$\displaystyle \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}$
• Feb 12th 2010, 09:57 AM
gary223
Quote:

Originally Posted by skeeter
$\displaystyle \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}$

Okay, I see. So whats the next step? Do I cancel some parts out or do I calculate each part? Or is that the question finished?
• Feb 12th 2010, 10:04 AM
skeeter
Quote:

Originally Posted by gary223
Okay, I see. So whats the next step? Do I cancel some parts out or do I calculate each part? Or is that the question finished?

substitute in the value of $\displaystyle \tan(45^\circ)$.
• Feb 12th 2010, 10:27 AM
gary223
Quote:

Originally Posted by skeeter
$\displaystyle \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}$

So its $\displaystyle \tan(\theta + 45^\circ)= \frac{tan\theta+1}{1-tan\theta.1}$

Correct?
• Feb 12th 2010, 10:30 AM
skeeter
Quote:

Originally Posted by gary223
So its $\displaystyle \tan(\theta + 45^\circ)= \frac{tan\theta+1}{1-tan\theta.1}$

Correct?

look again at what you were trying to prove ...
• Feb 12th 2010, 10:38 AM
gary223
Quote:

Originally Posted by skeeter
look again at what you were trying to prove ...

$\displaystyle \theta + 1= \frac{tan\theta+1}{1-tan\theta.1}$

???

EDIT:

Or is it:

$\displaystyle tan\theta + tan1= \frac{tan\theta+1}{1-tan\theta.1}$
• Feb 13th 2010, 06:42 AM
gary223
Or is it $\displaystyle tan\theta+ 1 = \frac{tan\theta+1}{1-tan\theta.1}$

Sorry for double post but I wanna get this.
• Feb 13th 2010, 06:59 AM
skeeter
you were asked to prove that $\displaystyle \tan(\theta+45^\circ) = \frac{1+\tan{\theta}}{1-\tan{\theta}}$

Krizalid gave you the sum identity for tangent ...

$\displaystyle \tan(x+y) = \frac{\tan{x} + \tan{y}}{1 - \tan{x} \cdot \tan{y}}$

my initial post basically told you to substitute $\displaystyle \theta$ for $\displaystyle x$ and $\displaystyle 45^\circ$ for $\displaystyle y$ ...

$\displaystyle \tan(\theta+45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}$

using the fact that $\displaystyle \tan(45^\circ) = 1$ ...

$\displaystyle \tan(\theta+45^\circ) = \frac{\tan{\theta} + 1}{1 - \tan{\theta}} = \frac{1+\tan{\theta}}{1 - \tan{\theta}}$

the original equation is proved ... you're done.
• Feb 13th 2010, 07:07 AM
gary223
Wow, it seems so easy once you do it. Thanks.
• Feb 13th 2010, 10:58 PM
Punch
Note that special angles exist, which are basically, sine, tangent and cosine of 30, 45, 60and 90. And you are required to recognise them..