Question

\(\sf cot(x\dfrac{\pi}{2})=-tan~x\)

Answer 1

what are we doin' here?

Answer 2

Whoops, proving it
\(\sf cot(x\dfrac{\pi}{2})=-tan~x\)

Answer 3

Looks there is a negative or positive sign missing on the left hand side inside cot

Answer 4

Gah, it should be \(\sf cot(x-\dfrac{\pi}{2})=-tan~x\)
sorry :/

Answer 5

\(\text{co}\) is actually defined such that \(\text{co}f(x) = f(\pi/2-x)\)

Answer 6

So we can start with: \[
\cot(x-\pi/2) = \tan(\pi/2-(x-\pi/2))
\]

Answer 7

Ok, that makes sense

Answer 8

I believe that simplifies to \(\tan(\pi-x)\).

Answer 9

One question, why is it tan instead of -tan?

Answer 10

The next step is probably the angle subtraction property.

Answer 11

angle difference property of tangent.

Answer 12

Hmm, but \(\tan(\pi)\) is undefined unfortunately...

Answer 13

Wait, no, \(\tan\pi = 0\).

Answer 14

\[
\tan(\pi-x)=\frac{\tan(\pi)-\tan(x)}{1+\tan(\pi)\tan(x)} = \frac{0-\tan(x)}{1+0} = -\tan(x)
\]