Prove tan

**gary223** · February 12th 2010, 09:24 AM

Prove that $tan(\theta+45degrees) = \frac{1+tan\theta}{1-tan\theta}$

**Krizalid** · February 12th 2010, 09:29 AM

straightforward, just use the formula for $\tan(x+y).$

**gary223** · February 12th 2010, 09:31 AM

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?

**Krizalid** · February 12th 2010, 09:32 AM

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

**gary223** · February 12th 2010, 09:38 AM

So what is x and y? Is it $1+tan\theta$ and $1-tan\theta$ ?

**skeeter** · February 12th 2010, 09:43 AM

Originally Posted by gary223

So what is x and y? Is it $1+tan\theta$ and $1-tan\theta$ ?

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

**gary223** · February 12th 2010, 09:57 AM

Originally Posted by skeeter

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

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?

**skeeter** · February 12th 2010, 10:04 AM

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 $\tan(45^\circ)$ .

**gary223** · February 12th 2010, 10:27 AM

Originally Posted by skeeter

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

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

Correct?

**skeeter** · February 12th 2010, 10:30 AM

Originally Posted by gary223

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

Correct?

look again at what you were trying to prove ...

**gary223** · February 12th 2010, 10:38 AM

Originally Posted by skeeter

look again at what you were trying to prove ...

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

???

EDIT:

Or is it:

$tan\theta + tan1= \frac{tan\theta+1}{1-tan\theta.1}$

**gary223** · February 13th 2010, 06:42 AM

Or is it $tan\theta+ 1 = \frac{tan\theta+1}{1-tan\theta.1}$

Sorry for double post but I wanna get this.

**skeeter** · February 13th 2010, 06:59 AM

you were asked to prove that $\tan(\theta+45^\circ) = \frac{1+\tan{\theta}}{1-\tan{\theta}}$

Krizalid gave you the sum identity for tangent ...

$\tan(x+y) = \frac{\tan{x} + \tan{y}}{1 - \tan{x} \cdot \tan{y}}<br />$

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

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

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

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

the original equation is proved ... you're done.

**gary223** · February 13th 2010, 07:07 AM

Wow, it seems so easy once you do it. Thanks.

**Punch** · February 13th 2010, 10:58 PM

Note that special angles exist, which are basically, sine, tangent and cosine of 30, 45, 60and 90. And you are required to recognise them..

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