$\tan(x)=\cot(90^\circ-x)$??

Question

I was looking at a mark scheme for a question I was stuck on, and I came across this. You are asked to work out the value of $\tan 75^\circ$ after you've worked out $\cos 15^\circ$ and $\sin 15^\circ$. I noticed that $\tan(x)=\cot(90^\circ-x)$. I've never seen this before, and this makes no sense to me, so please could someone explain it to me? Are there any other similar trig properties that I should know about?

Answer 1

so $\tan\theta = \frac{a}{b} = \cot(90-\theta)$.

By the way your computation above computes $\cot 75$, not $\tan 75$.

Answer 2

You might want to learn here about trigonometric identities. Note that, $\cot(90^\circ-x)$ means $x$ is reflected from angle $90^\circ$ and in this case, the result is $\tan x$. Just click the given link. I hope this helps.

Answer 3

This follows from simple well-known trigonometric identities:

$$\tan(90-x) = \frac{\sin(90-x)}{\cos(90-x)}=\frac{\cos(x)}{\sin(x)}=\cot(x)$$