Question Number 108950 by ZiYangLee last updated on 20/Aug/20
$$\mathrm{In}\:\bigtriangleup\mathrm{ABC},\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}+\mathrm{cosC}}{\mathrm{1}+\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosC}}=\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{cot}\frac{\mathrm{C}}{\mathrm{2}} \\ $$
Answered by mnjuly1970 last updated on 20/Aug/20
![(((1+cos(C))+(cos(A)−cos(B)))/((1−cos(C))+(cos(A)+cos(B)))) =((2cos^2 ((C/2))−2sin(((A+B)/2))sin(((A−B)/2)))/(2sin^2 ((C/2))+2cos(((A+B)/2))cos(((A−B)/2)))) =((2cos((C/2))[sin(((A+B)/2))−sin(((A−B)/2))])/(2sin((C/2))[cos(((A+B)/2)+co(((A−B)/2))])) =((4cos((C/2))cos((A/2))sin((B/2)))/(4sin((C/2))cos((A/2))cos((B/2))))=cot((C/2))tan((B/2))...▲ .....M.N....](https://www.tinkutara.com/question/Q108964.png)
$$\frac{\left(\mathrm{1}+{cos}\left(\mathrm{C}\right)\right)+\left({cos}\left(\mathrm{A}\right)−{cos}\left(\mathrm{B}\right)\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{C}\right)\right)+\left({cos}\left(\mathrm{A}\right)+{cos}\left(\mathrm{B}\right)\right)} \\ $$$$=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{C}}{\mathrm{2}}\right)−\mathrm{2}{sin}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{C}}{\mathrm{2}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}{cos}\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\left[{sin}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)−{sin}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)\right]}{\mathrm{2}{sin}\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\left[{cos}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}+{co}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)\right]\right.} \\ $$$$=\frac{\mathrm{4}{cos}\left(\frac{\mathrm{C}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{A}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{4}{sin}\left(\frac{\mathrm{C}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{A}}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{B}}{\mathrm{2}}\right)}={cot}\left(\frac{\mathrm{C}}{\mathrm{2}}\right){tan}\left(\frac{\mathrm{B}}{\mathrm{2}}\right)…\blacktriangle \\ $$$$\:\:\:\:\:\:\:\:\:…..\mathscr{M}.\mathscr{N}…. \\ $$