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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

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    $\begingroup$ 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. $\endgroup$ – DonAntonio Apr 19 '14 at 15:56
  • $\begingroup$ @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 $\endgroup$ – Jim Apr 19 '14 at 16:12
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    $\begingroup$ @Jim use degrees while plotting $\endgroup$ – evil999man Apr 19 '14 at 16:17
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    $\begingroup$ @Jim, weren't you taught the definitions of sine and cosine using a straight-angled triangle?? $\endgroup$ – DonAntonio Apr 19 '14 at 16:17
  • $\begingroup$ @Awesome I'm pretty sure I set my software to degrees, but I'll double check $\endgroup$ – Jim Apr 19 '14 at 16:21
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enter image description here

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

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

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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.

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This follows from simple well-known trigonometric identities:

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

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