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Question Number 43810 by gyugfeet last updated on 15/Sep/18

((1−cosθ+coβ−cos(θ+β))/(1+cosθ−cosβ−cos(θ+β)))=tan(θ/2). cot (β/2)

$$\frac{\mathrm{1}−{cos}\theta+{co}\beta−{cos}\left(\theta+\beta\right)}{\mathrm{1}+{cos}\theta−{cos}\beta−{cos}\left(\theta+\beta\right)}={tan}\frac{\theta}{\mathrm{2}}.\:{cot}\:\frac{\beta}{\mathrm{2}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18

=((2sin^2 (θ/2)+2sin((θ/2)+β)sin((θ/2)))/(2cos^2 ((θ/2))−2cos((θ/2)+β)cos((θ/2)))) =tan(θ/2)×((sin(θ/2)+sin((θ/2)+β))/(cos((θ/2))−cos((θ/2)+β))) =tan((θ/2))×((2sin(((θ+β)/2))cos((β/2)))/(2sin(((θ+β)/2))sin((β/2)))) =tan((θ/2))cot((β/2))

$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right){sin}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$={tan}\frac{\theta}{\mathrm{2}}×\frac{{sin}\frac{\theta}{\mathrm{2}}+{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)−{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right)} \\ $$$$={tan}\left(\frac{\theta}{\mathrm{2}}\right)×\frac{\mathrm{2}{sin}\left(\frac{\theta+\beta}{\mathrm{2}}\right){cos}\left(\frac{\beta}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\theta+\beta}{\mathrm{2}}\right){sin}\left(\frac{\beta}{\mathrm{2}}\right)} \\ $$$$={tan}\left(\frac{\theta}{\mathrm{2}}\right){cot}\left(\frac{\beta}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$

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