Question Number 201425 by mathlove last updated on 06/Dec/23
$${prove}\:{that} \\ $$$$\frac{\mathrm{1}−{cosA}+{cosB}−{cos}\left({A}+{B}\right)}{\mathrm{1}+{cosA}−{cosB}−{cos}\left({A}+{B}\right)}={tan}\frac{{A}}{\mathrm{2}}\centerdot{cot}\frac{{B}}{\mathrm{2}} \\ $$
Answered by esmaeil last updated on 06/Dec/23
$$\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{A}+{B}}{\mathrm{2}}\right)\:−\mathrm{2}{sin}\left(\frac{{B}+{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}−{A}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{B}+{A}}{\mathrm{2}}\right)−\mathrm{2}{sin}\left(\frac{{B}+{A}}{\mathrm{2}}\right){sin}\left(\frac{{A}−{B}}{\mathrm{2}}\right)}= \\ $$$$\frac{{sin}\frac{{B}+{A}}{\mathrm{2}}+{sin}\left(\frac{{A}−{B}}{\mathrm{2}}\right)}{{sin}\left(\frac{{B}+{A}}{\mathrm{2}}\right)−{sin}\left(\frac{{A}−{B}}{\mathrm{2}}\right)}\:=\frac{\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{B}}{\mathrm{2}}}{\mathrm{2}{sin}\frac{{B}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}}} \\ $$$${tan}\frac{{A}}{\mathrm{2}}{cot}\frac{{B}}{\mathrm{2}}\:\:\left(\mathrm{1}−{cos}\left({AB}+{A}\right),{cosB}−{cosA}\right). \\ $$