If tan x

Given, $\quad \tan x=\frac{1-\cos y}{\sin y}$

$\left[\right.$ using identity, $\left.\quad \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right]$

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

$=\frac{2\left(\frac{1-\cos y}{\sin y}\right)}{1-\left(\frac{1-\cos y}{\sin y}\right)^{2}}=\frac{2\left(\frac{2 \sin ^{2} y / 2}{2 \sin ^{y} / 2 \cos y / 2}\right)}{1-\left(\frac{2 \sin ^{2} y / 2}{2 \sin y / 2 \cos y / 2}\right)^{2}}$

using identity : $1-\cos \theta=2 \sin ^{2} \theta / 2$

and $\quad \sin \theta=2 \sin \theta / 2 \cos \theta / 2$

$=\frac{2\left(\frac{\sin ^{y} / 2}{\cos ^{y} / 2}\right)}{1-\left(\frac{\sin ^{y} / 2}{\cos ^{y} / 2}\right)^{2}}$

$=\frac{2 \tan ^{y} / 2}{1-\tan ^{2} y / 2}$

$=\operatorname{tany} \quad\left(\right.$ using identity $\left.: \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\tan 2 \theta\right)$

i. e. $\quad \tan 2 \mathrm{x}=\operatorname{tany}$