Found an odd relationship! Could someone help me to prove or debunk it? I finished up in hospital which typically means that one has A LOT of spare time to kill and after using electronic devices so much that it makes you sorry I flinched into doodling and and light-headedly playing with a calculator.
It happend than that I bounced upon one odd thing:
For every right-angled triangle I sketched it seemed to be true that
$$\frac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{\cos^2{\alpha}+\cos^2{\beta}+\cos^2{\gamma}}=2$$
Because my medicament dispenser is kinda slacking my brain I can't really concentrate on finding a way to prove this relation and a counterexample is also not in sight. So can someone please help me out? This open question is always on my mind and starts to become annonying :)
yours, Levix
 A: In a right triangle, exactly one of $\alpha$, $\beta$, $\gamma$ must equal $\pi/2$, so let this be $\gamma = \pi/2$.  Then $\sin \gamma = 1$ and $\cos \gamma = 0$.  We must also have $\alpha + \beta = \pi/2$, hence $\beta = \pi/2 - \alpha$ and it immediately follows that $\sin \beta = \sin (\pi/2 - \alpha) = \cos \alpha$, and $\cos \beta = \sin \alpha$.  The rest is simple substitution and the circular identity $\sin^2 \alpha + \cos^2 \alpha = 1$.
A: L.H.S.
$$\large\frac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{\cos^2{\alpha}+\cos^2{\beta}+\cos^2{\gamma}}$$
Since, 
$\alpha+\beta+\gamma$=$180^\circ$
Since, the triangle is a right angled triangle (assuming it is at $\gamma$), that means
$\alpha+\beta=90^\circ$
Or, 
$\beta=90^\circ-\alpha$
So, 
$\sin \beta$=$\cos \alpha$
and 
$\cos \beta$=$\sin \alpha$
The equation then changes to :
$$\large\frac{\sin^2 \alpha+\cos^2 \alpha+1}{\cos^2 \alpha+\sin^2 \alpha+ 0}$$
$$\large\frac{1+1}{1 + 0}=2=\text {R.H.S}$$
A: WLOG $\displaystyle\alpha=90^\circ\implies\beta+\gamma=90^\circ\iff\gamma=90^\circ-\beta$
$$\sin^2\beta+\sin^2\gamma=\sin^2\beta+\sin^2(90^\circ-\beta)=\sin^2\beta+\cos^2\beta=?$$
$$\cos^2\beta+\cos^2\gamma=?$$
A: $cos^2(\alpha)+cos^2(\beta)+cos^2(\gamma)=t =>$$ sin^2(\alpha)+sin^2(\beta)+sin^2(\gamma)=1-cos^2(\alpha)+1-cos^2(\beta)+1-cos^2(\gamma)=3-t$
where $-3<=t<=3$
then $\frac{3-t}{t}=2 => 2t=3-t => t=1$
that means $cos^2(\frac\pi2)+cos^2(\beta)+cos^2(\frac\pi2-\beta)=1$
that means $0+cos^2(\beta)+sin^2(\beta)=1$
and it is always true(does not matter which angle is $\frac\pi2$ and $cos(\alpha)=sin(\frac\pi2-\alpha) is always true $ )
or directly going $ \frac{1+sin^2(\alpha)+cos^2(\alpha)}{0+sin^2(\alpha)+cos^2(\alpha)}= \frac{1+1}{0+1}=2$ 
