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

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- Feb 12th 2010, 09:24 AMgary223Prove tan
Prove that $\displaystyle tan(\theta+45degrees) = \frac{1+tan\theta}{1-tan\theta}$

- Feb 12th 2010, 09:29 AMKrizalid
straightforward, just use the formula for $\displaystyle \tan(x+y).$

- Feb 12th 2010, 09:31 AMgary223
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 AMKrizalid
$\displaystyle \tan(x+y)=\frac{\tan x+\tan y}{1-\tan(x)\tan(y)}.$

- Feb 12th 2010, 09:38 AMgary223
So what is x and y? Is it $\displaystyle 1+tan\theta$ and $\displaystyle 1-tan\theta$?

- Feb 12th 2010, 09:43 AMskeeter
- Feb 12th 2010, 09:57 AMgary223
- Feb 12th 2010, 10:04 AMskeeter
- Feb 12th 2010, 10:27 AMgary223
- Feb 12th 2010, 10:30 AMskeeter
- Feb 12th 2010, 10:38 AMgary223
- Feb 13th 2010, 06:42 AMgary223
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 AMskeeter
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 AMgary223
Wow, it seems so easy once you do it. Thanks.

- Feb 13th 2010, 10:58 PMPunch
Note that special angles exist, which are basically, sine, tangent and cosine of 30, 45, 60and 90. And you are required to recognise them..