Trigonometry / Geometry Puzzle with a Circle Inscribed within a Square

Point P is any point on the inscribed circle. You must prove that

(tan(a))^2 + (tan(B))^2 = 8

I first moved point P down to the point where the square would be tangent to the curve to make the problem easier. I realized that (tan(a))^2 = (tan(b))^2 = 4 in this simplified case.

Next I separated (tan(a))^2 into

sin(a) /cos(a) *sin(a)/cos(a) =4


and from there

sin(a)/cos(a) =2


Then I used the law of sine to find that

[sin(a)/(2^(1/2)*AB)] = [sin(A)/(1/2AB)]

And that sin(a) = 2*2^(1/2)*sin(A)

After that I was kind of lost on what to do.

Any help would be appreciated! Thank you in advance!

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Typo: in the simplified case, you mean $=8$, not $=4$. – André Nicolas Mar 14 '13 at 3:29
Please be careful with capitalization: $\tan(B)$ is not the same as $\tan(b)$ and is meaningless as $B$ is a point. – Ross Millikan Mar 14 '13 at 3:39
If $P$ is not on $CD$, it's not clear to me what angles $a$ and $b$ are. Is angle $a$ supposed to be angle $APC$ in that case? – Gerry Myerson Mar 14 '13 at 5:38
Yes, angle a is to be angle APC and angle b angle DPB. Sorry for the ambiguity! – Hannah Mar 15 '13 at 0:49

There is a significant ambiguity in the problem statement. Are we to prove that the statement is true for any $P$ on the circle, or are we given that $P$ is on the circle and we have to prove the result? In the second case, the question implies that the value is independent of which $P$ is chosen on the circle, so we can choose one that is convenient. In the second case, we rely on the problem giver to have shown that the result doesn't matter which $P$ is chosen, so we can choose our favorite. In that case, I choose $P$ as the midpoint of the bottom side of the square. Then for a unit square $PA=PB=\frac {\sqrt 5}2$ so $\sin a=\sin b=-\frac 2{\sqrt 5}, \cos a=\cos b = \frac 1{\sqrt 5}, \tan a = \tan b = -2$ so $(\tan a)^2+ (\tan b)^2=8$
My interpretation is that OP knows it's true for that particular $P$ (even though OP typoed the $8$ into a $4$) and wants a proof that it's true for all $P$ on the circle. – Gerry Myerson Mar 14 '13 at 4:28