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

Solution 1:

WLOG $\displaystyle\alpha=90^\circ\implies\beta+\gamma=90^\circ\iff\gamma=90^\circ-\beta$

$$\sin^2\beta+\sin^2\gamma=\sin^2\beta+\sin^2(90^\circ-\beta)=\sin^2\beta+\cos^2\beta=?$$

$$\cos^2\beta+\cos^2\gamma=?$$

Solution 2:

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 $\sin^2 \alpha + \cos^2 \alpha = 1$.