1. ## Prove tan

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

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

3. 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?

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

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

6. Originally Posted by gary223
So what is x and y? Is it $\displaystyle 1+tan\theta$ and $\displaystyle 1-tan\theta$?
$\displaystyle \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}$

7. 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?

8. 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 $\displaystyle \tan(45^\circ)$.

9. 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?

10. 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 ...

11. 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:

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

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

Sorry for double post but I wanna get this.

13. 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.

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

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