Question Number 43810 by gyugfeet last updated on 15/Sep/18
$$\frac{\mathrm{1}−{cos}\theta+{co}\beta−{cos}\left(\theta+\beta\right)}{\mathrm{1}+{cos}\theta−{cos}\beta−{cos}\left(\theta+\beta\right)}={tan}\frac{\theta}{\mathrm{2}}.\:{cot}\:\frac{\beta}{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18
$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right){sin}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$={tan}\frac{\theta}{\mathrm{2}}×\frac{{sin}\frac{\theta}{\mathrm{2}}+{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)−{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right)} \\ $$$$={tan}\left(\frac{\theta}{\mathrm{2}}\right)×\frac{\mathrm{2}{sin}\left(\frac{\theta+\beta}{\mathrm{2}}\right){cos}\left(\frac{\beta}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\theta+\beta}{\mathrm{2}}\right){sin}\left(\frac{\beta}{\mathrm{2}}\right)} \\ $$$$={tan}\left(\frac{\theta}{\mathrm{2}}\right){cot}\left(\frac{\beta}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$