Here's a slightly non-obvious relation between some angles, that I ran into accidentally.
Claim: (angles below given in degrees)
\[ \tan(10) \tan(50) = \tan(20) \tan(30) \]
Verification
Claim is equivalent to
\[ \sin(10)\sin(50) \cos(20) \cos(30) - \sin(20)\sin(30) \cos(10) \cos(50) = 0 \]
the left hand side is equal to (by angle addition formula)
\[ \frac12 \left[ \cos(30 + 20) \cos(50-10) - \cos(30-20) \cos(50+10) \right] \]
so the claim is equivalent to
\[ \cos(50) \cos(40) - \cos(10) \cos(60) = 0 \]
Observe that $\cos(50) = \sin(40)$ and vice versa, so we can write
\[ \cos(50) \cos(40) = \frac12 \left[ \cos(50) \cos(40) + \sin(50) \sin(40) \right] = \frac12 \cos(50 - 40) \]
again using angle addition, and finally since $\cos(60) = \frac12$, the claim is proved.
Generalization for any $\gamma$ (measured in degrees)
\[ \tan(\gamma - 15) \tan(75-\gamma) = \tan(45-\gamma) \tan(3\gamma - 45) \]
our original claim happens to be the case $\gamma = 25$.
