Does $\cot^2 $ equal $ \cos^2/ \sin^2$?

Question

In pre-calc, we're doing trigonometric identities, and I have a question but my teacher isn't available right now. We have to verify trigonometric identities, but I can't progress right now because I'm unsure if cotangent squared equals cosine squared over sine squared. I'm not asking for the answer to my problem, I just don't know if my thinking pattern is correct here.

Answer 1

$$\cot^2x=(\cot x)^2=\left(\dfrac{\cos x}{\sin x}\right)^2=\dfrac{\cos^2x}{\sin^2x}$$

Answer 2

Yes, by definition $\cot x = \frac{1}{\tan x}$, and $\tan x = \frac{\sin x}{\cos x}$, so $$\cot x = \frac{1}{\sin x/\cos x} = \frac{\cos x}{\sin x}.$$ Thus, squaring both sides of the equation, we obtain $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$.

Answer 3

Since the cotangent function is defined as cosine divided by sine $($or as reciprocal tangent, which is defined by sine divided by cosine, respectively$)$ we get that

$$\cot^2(x)=(\cot(x))^2=\left(\frac{\cos(x)}{\sin(x)}\right)^2=\frac{(\cos(x))^2}{(\sin(x))^2}=\frac{\cos^2(x)}{\sin^2(x)}$$

$$\therefore~\cot^2(x)~=~\frac{\cos^2(x)}{\sin^2(x)}$$

Answer 4

$$ \frac{\sin(x)}{\cos(x)} = \tan(x) $$

Thus $$ \frac{\cos(x)}{\sin(x)} = \frac{1}{\tan(x)} = \cot(x) $$ and $$ \Big(\frac{\cos(x)}{\sin(x)}\Big)^{2} = \frac{\cos^{2}(x)}{\sin^{2}(x)} = \cot^{2}(x) $$