Prove $\cos^2(\theta)+\sin^2(\theta) = 1$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$
I solved this by using right triangle,
$$\sin(\theta) = \frac{a}{c}, \quad \cos(\theta) = \frac{b}{c}$$
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
$$\Bigl(\frac{b}{c}\Bigr)^2 + \Bigl(\frac{a}{c}\Bigr)^2 = 1 $$
$$\frac {a^2 + b^2} {c^2} = 1 $$
now using Pythagorean identity: $a^2 + b^2 = c^2$
$$\frac {c^2} {c^2} = 1, \quad 1 = 1 $$

You solved this by using some right triangle but there isn't a right triangle in your solution? Remember, things like that have to be explicitly written in your proof.
Also, the proof is not very valid because you started off by asserting that:
$$\sin^2(\theta)+\cos^2(\theta) = 1$$
when really, what you should have started with is just the left-hand side of that. In other words, I can  see where you're going with the reasoning but it isn't airtight because of your wording.
Really, what you should have done is to say that:
$$\sin(\theta) = \frac{a}{c}$$
$$\cos(\theta) = \frac{b}{c}$$
where $a,b,c$ are the sides of a particular right triangle whose image you're supposed to include. Then, you have:
$$\sin^2(\theta)+\cos^2(\theta) = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2}$$
By Pythagoras's Theorem relating the sides of the right triangle, it is the case that $a^2+b^2 = c^2$. Hence, we have:
$$\sin^2(\theta) + \cos^2(\theta) = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1$$
which proves the desired result.
