The Pythagorean identity is just Pythagoras, divided

There are four identities every trigonometry course puts on the first page:

$$\sin^2\alpha + \cos^2\alpha = 1$$ $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} \qquad \cot\alpha = \frac{\cos\alpha}{\sin\alpha} \qquad \tan\alpha \cdot \cot\alpha = 1$$

None of them is a new fact. Each is a two-line consequence of the definitions, and the first one is a theorem you have known since primary school wearing a different hat.

Two lines

Work in a right triangle with legs $a$ (opposite $\alpha$) and $b$ (adjacent), and hypotenuse $c$. By definition $\sin\alpha = \frac{a}{c}$ and $\cos\alpha = \frac{b}{c}$. Square both and add:

$$\sin^2\alpha + \cos^2\alpha = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1$$

The second-to-last step is Pythagoras — $a^2 + b^2 = c^2$ — and nothing else.

So the Pythagorean identity really is Pythagoras, divided through by $c^2$. That is worth holding on to, because it means you can never truly forget it: if the identity slips your mind mid-problem, draw the triangle and divide.

Divide again, get two more

The interesting part is what happens when you divide the identity by something other than $1$. Take $\sin^2\alpha + \cos^2\alpha = 1$ and divide through by $\cos^2\alpha$:

$$\tan^2\alpha + 1 = \frac{1}{\cos^2\alpha} \qquad\Longrightarrow\qquad \cos^2\alpha = \frac{1}{1 + \tan^2\alpha}$$

Now substitute that back into $\sin^2\alpha = 1 - \cos^2\alpha$:

$$\sin^2\alpha = 1 - \frac{1}{\tan^2\alpha+1} = \frac{\tan^2\alpha + 1 - 1}{\tan^2\alpha+1} = \frac{\tan^2\alpha}{1+\tan^2\alpha}$$

Which leaves you with the pair:

$$\sin^2\alpha = \frac{\tan^2\alpha}{1+\tan^2\alpha} \qquad \cos^2\alpha = \frac{1}{1+\tan^2\alpha}$$

Why these two are worth memorising

These are what you reach for when a problem hands you $\tan\alpha$ and asks for the other functions. The instinct is to recover the angle first — find $\alpha$ from $\tan\alpha$, then take its sine. Without a calculator that is simply not possible, and with one it throws away exactness.

These two identities go straight from $\tan\alpha$ to $\sin^2\alpha$ and $\cos^2\alpha$ with no angle in sight. The only thing left to decide is the sign when you take the square root, and that comes from the quadrant $\alpha$ lives in — never from the identity itself.

That last point is where most marks are lost. The identity gives you the magnitude; the quadrant gives you the sign. A problem that says ”$\tan\alpha = 2$ and $\alpha$ is in the third quadrant” is telling you the second half on purpose, and an answer that ignores it is half an answer.