Small question on Pythagoras theorem The projections-of-the-legs over the hypotenuse should add up to the hypotenuse $c$.
Is there any alternative way to prove below?
$$a\cos \alpha + b\sin \alpha = \sqrt{a^2+b^2}$$


Well, $$\cos \alpha = \frac ac\quad \&\quad \sin \alpha = \frac bc$$
so $$a\cos \alpha +b\sin \alpha = \frac 1c \times (a^2+b^2)=\frac {c^2}c=c$$
Of course, that last step requires the Pythagorean Theorem ("PT").  It is worth remarking that, without using PT the argument shows $$a\cos \alpha +b\sin \alpha = \frac {a^2+b^2}c$$ and since the OP has shown that $$a\cos \alpha +b\sin \alpha = c$$ we can combine the two arguments to get $$\frac {a^2+b^2}c=c \implies a^2+b^2=c^2$$ so the two arguments together yield a proof of PT.
