Question

If $\tan x=\frac{1-\cos y}{\sin y},$ then tan 2x = _____________.

Answer 1

$$\begin{gathered} \mathrm{Given},\tan x=\frac{1-\cos y}{\sin y} \\ \left[\mathrm{using}\mathrm{identity},\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\right] \\ \tan 2x=\frac{2\tan x}{1-{\tan }^{2}x} \\ =\frac{2\left({\displaystyle \frac{1-\cos y}{\sin y}}\right)}{1-{\left({\displaystyle \frac{1-\cos y}{\sin y}}\right)}^2}=\frac{2\left({\displaystyle \frac{2{\sin }^2{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2\sin {\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2\cos {\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$}\right.}}}\right)}{1-{\left({\displaystyle \frac{2{\sin }^2{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2\sin {\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos {\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)}^2} \\ \mathrm{using}\mathrm{identity}:1-\mathrm{cos\theta }=2{\sin }^2\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{and}\mathrm{sin\theta }=2\sin \raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\cos \raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ =\frac{2\left({\displaystyle \frac{\sin {\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\cos {\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)}{1-{\left({\displaystyle \frac{\sin {\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\cos {\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)}^2} \\ =\frac{2\tan {\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1-{\tan }^2{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ =\mathrm{tany}\left(\mathrm{using}\mathrm{identity}:\frac{2\mathrm{tan\theta }}{1-{\tan }^2\mathrm{\theta }}=\tan 2\mathrm{\theta }\right) \\ \mathrm{i}.\mathrm{e}.\tan 2\mathrm{x}=\mathrm{tany} \end{gathered}$$