Continued Fraction expansion of $\tan(1)$ Prove that the continued fraction of $\tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,...]$.  I tried using the same sort of trick used for finding continued fractions of quadratic irrationals and trying to find a recurrence relation, but that didn't seem to work.

We  use the formula given here: Gauss' continued fraction for $\tan z$ and see that
$$\tan(1) = \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5 -\dots}}}$$
Now use the identity
$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$
To transform $$\cfrac{1}{a - \cfrac{1}{b - \cfrac{1}{c - \dots}}}$$ to
$$\cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-2 + \dots}}}}}$$
to get the expansion for $\displaystyle \tan(1)$
The above expansion for $\tan(1)$ becomes
$$ \cfrac{1}{1-1 + \cfrac{1}{1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}}}$$
$$  = 1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}$$
$$= 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{5 + \dots}}}}}$$
To prove the transformation, 
let $\displaystyle x = b - \cfrac{1}{c - \dots}$
Then
$$ \cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$
$$ = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-1 + \cfrac{1}{c - \dots}}}}$$
Applying the identity again to 
$$\cfrac{1}{b-1 + \cfrac{1}{c - \dots}}$$
we see that
$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-1 + \cfrac{1}{d - \dots}}}}}}$$
Applying again to $\cfrac{1}{c-1 + \cfrac{1}{d - \dots}}$ etc gives the required CF.
