If $\alpha<45^\circ$, then $\tan\alpha<\cot\alpha$ using unit circe I am trying to show (prove) using the diagram below of the unit circle that if $\alpha<45^\circ$, then $\tan\alpha<\cot\alpha.$

It all comes down to showing $AP=\tan\alpha<OQ_1=\cot\alpha.$ We can note that since $\alpha<45^\circ,$ then all of its trig functions are positive and we can say they're equal to the respective segments (as I have denoted on the diagram).
Also I would like to ask you what happens if $\alpha$ is a negative angle as the text of the problem doesn't expressly states that it cannot be such? As a matter of fact on the diagram in my book the direction of the angle alpha is from OA to OM (a directed arrow). I am not sure this means it's positive.


If $\alpha=45^\circ$ then triangle $OAP$ is isosceles. As $\alpha < 45^\circ$, we see that $AP$ is strictly shorter than $OA$.
(This comes just from visual intuition, but it is equivalent to the fact that $\tan\alpha<1$ for $\alpha\in [0^\circ,45^\circ)$. It essentially follows from the definition of degrees, which tells us that a line at a $45^\circ$ angle to the horizontal has gradient $1$, and lines at smaller positive angles have gradients in $[0,1)$.)
Then, we must have that $\tan \alpha=AP<OA\le OQ_1=\cot\alpha$, which proves the statement in question. Notice that $OA=1$ so we've also proven $\tan \alpha<1<\cot\alpha$.
By noticing that $\tan$ and $\cot$ are both odd functions, or simply that they are negative in the range $(-45^\circ,0^\circ)$, we see that $\cot\alpha<\tan\alpha$ in the case that $\alpha\in (-45^\circ,0^\circ)$, and $\cot$ is undefined at $0^\circ$, so I think you are supposed to make the implicit assumption that $\alpha$ is positive.
