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