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

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?

• Have you seen $\;\sin(90-x)=\cos x\;,\;\;\cos(90-x)=\sin x\;$ ? This follows at once from the most basic definition of trigonometric functions in high school. – DonAntonio Apr 19 '14 at 15:56
• @DonAntonio I've never seen those before, (they're probably not on my syllabus then), but I can see how they work, since you're just translating the graph, right? I went on graphing software and entered sin(90-x) and cos(x), but the sin(90-x) graph was not translated enough to become the cos(x) graph. I think I went wrong somewhere then – Jim Apr 19 '14 at 16:12
• @Jim use degrees while plotting – evil999man Apr 19 '14 at 16:17
• @Jim, weren't you taught the definitions of sine and cosine using a straight-angled triangle?? – DonAntonio Apr 19 '14 at 16:17
• @Awesome I'm pretty sure I set my software to degrees, but I'll double check – Jim Apr 19 '14 at 16:21

so $\tan\theta = \frac{a}{b} = \cot(90-\theta)$.
By the way your computation above computes $\cot 75$, not $\tan 75$.
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.
$$\tan(90-x) = \frac{\sin(90-x)}{\cos(90-x)}=\frac{\cos(x)}{\sin(x)}=\cot(x)$$