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

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