1. ## Trig identities!!!!!!!!!

if cotx= -2/3 and pi/2<x<pi

solve for tan(x/2)

Thanks

2. $\displaystyle \frac{\pi}{2}<x<\pi\;\implies\;\frac{\pi}{4}<\frac {x}{2}<\frac{\pi}{2}$

3. Hello, andymac!

If $\displaystyle \cot x\:=\: -\frac{2}{3}\:\text{ and }\;\frac{\pi}{2} < x < \pi,\:\text{ find: }\tan \frac{x}{2}$

$\displaystyle \cot x \:=\:-\frac{2}{3}\quad\Rightarrow\quad \tan x \:=\:-\frac{3}{2} \:=\:\frac{opp}{adj}$

Since $\displaystyle x$ is in Quadrant 2, $\displaystyle opp = 3,\:adj = -2 \quad\Rightarrow\quad hyp = \sqrt{13}$
. . Hence: .$\displaystyle \cos x \:=\:-\frac{2}{\sqrt{13}}$

Half-angle identity: .$\displaystyle \tan\frac{x}{2} \;=\;\pm\sqrt{\frac{1-\cos x}{1 + \cos x}}$

Since $\displaystyle x$ is in Quadrant 2, $\displaystyle \frac{x}{2}$ is in Quadrant 1.

We have: .$\displaystyle \tan\frac{x}{2} \;=\;\sqrt{\frac{1-(-\frac{2}{\sqrt{13}})}{1 + (-\frac{2}{\sqrt{13}})}} \;=\;\sqrt{\frac{\sqrt{13} + 2}{\sqrt{13} - 2}}$

Rationalize: .$\displaystyle \tan\frac{x}{2} \;= \;\sqrt{\frac{\sqrt{13}+2}{\sqrt{13}-2}\cdot{\color{blue}\frac{\sqrt{13}+2}{\sqrt{13}+2 }}} \;=\;\sqrt{\frac{(\sqrt{13} + 2)^2}{9}}$

Therefore: .$\displaystyle \boxed{\tan\frac{x}{2} \;=\;\frac{\sqrt{13}+2}{3}}$