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

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Answer 1 · 2018-05-20T07:40:55

$tan x = \frac{2 k}{(1 + k) (1 - k)}$ $color(maroon)(tan x = (2k)/((1+k)(1-k))$

$cot (\frac{x}{2}) = - k, tan x = ?$ $cot (x/2) = -k, tan x = ?$

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

$tan x = \frac{2 tan (\frac{x}{2})}{1 - {tan}^{2} (\frac{x}{2})}$ $color(green)(tan x = (2 tan(x/2)) / (1 - tan^2 (x/2))$ , identity

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

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

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

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

$tan x = \frac{2 k}{(1 + k) (1 - k)}$ $color(maroon)(tan x = (2k)/((1+k)(1-k))$

Answer 2 · 2018-05-20T07:42:21

$tan x = \frac{- 2 k}{k^{2} - 1}$ $tanx=(-2k)/(k^2-1)$

We know that,

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

$cot(x/2)=-k=>tan(x/2)=-1/k...to(2)$

Using $(1)$

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

$=>tanx=(2(-1/k))/(1-(-1/k)^2)...toApply(2)$

$=>tanx=(-2/k)/(1-1/(k^2))$

$=>tanx=(-2/k)/((k^2-1)/k^2)$

$=>tanx=(-2k)/(k^2-1)$