# Cot x/2=-k then tan x=?

May 20, 2018

color(maroon)(tan x = (2k)/((1+k)(1-k))

#### Explanation:

cot (x/2) = -k, tan x = ?

$\tan \left(\frac{x}{2}\right) = - \left(\frac{1}{k}\right)$

color(green)(tan x = (2 tan(x/2)) / (1 - tan^2 (x/2)), identity

Substituting for $\tan \left(\frac{x}{2}\right) = - \frac{1}{k}$,

$\tan x = \frac{- \frac{2}{k}}{1 - {\left(- \frac{1}{k}\right)}^{2}}$

$\tan x = \frac{\frac{2}{k}}{\left(\frac{1}{k} ^ 2\right) - 1}$

$\tan x = \frac{\frac{2}{k}}{\frac{1 - {k}^{2}}{k} ^ 2}$

color(maroon)(tan x = (2k)/((1+k)(1-k))

May 20, 2018

$\tan x = \frac{- 2 k}{{k}^{2} - 1}$

#### Explanation:

We know that,

color(blue)(tantheta=(2tan(theta/2))/(1-tan^2(theta/2))...to(1) [half angle formula]
Now,given that,

$\cot \left(\frac{x}{2}\right) = - k \implies \tan \left(\frac{x}{2}\right) = - \frac{1}{k} \ldots \to \left(2\right)$

Using $\left(1\right)$

tanx=(2tan(x/2))/(1-tan^2(x/2)

$\implies \tan x = \frac{2 \left(- \frac{1}{k}\right)}{1 - {\left(- \frac{1}{k}\right)}^{2}} \ldots \to A p p l y \left(2\right)$

$\implies \tan x = \frac{- \frac{2}{k}}{1 - \frac{1}{{k}^{2}}}$

$\implies \tan x = \frac{- \frac{2}{k}}{\frac{{k}^{2} - 1}{k} ^ 2}$

$\implies \tan x = \frac{- 2 k}{{k}^{2} - 1}$