TITLE: Found an odd relationship! Could someone help me to prove or debunk it?
QUESTION [18 upvotes]: 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

REPLY [4 votes]: 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 :
$$\large\frac{\sin^2 \alpha+\cos^2 \alpha+1}{\cos^2 \alpha+\sin^2 \alpha+ 0}$$
$$\large\frac{1+1}{1 + 0}=2=\text {R.H.S}$$