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

trigonometry

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?

enter image description here

Best Answer

enter image description here

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

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