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Question-94523

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Question Number 94523 by peter frank last updated on 19/May/20
Commented by PRITHWISH SEN 2 last updated on 19/May/20
By Napier′s Analogy for any△ABC tan((B−C)/2) =((b−c)/(b+c)) cot(A/2) −tan((A/2)+B)= ((b+c)/(b−c)) .tan(A/2) ((tan(A/2)cotB+1)/(tan(A/2)−cotB)) = ((b+c)/(b−c)) tan(A/2) tan(A/2)cotB +1 =((b+c)/(b−c))(tan^2 (A/2)−tan (A/2)cot B) tan(A/2)cotB(1+((b+c)/(b−c)))=((b+c)/(b−c)) tan^2 (A/2) −1 ((2b)/(b−c))cotB=((b+c)/(b−c)) tan (A/2)−cot(A/2) ∴ (b+c)tan(A/2)−(b−c)cot(A/2) = 2bcotB Hence proved
$$\mathrm{By}\:\mathrm{Napier}'\mathrm{s}\:\mathrm{Analogy}\: \\ $$$$\mathrm{for}\:\mathrm{any}\bigtriangleup\mathrm{ABC} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{B}}−\boldsymbol{\mathrm{C}}}{\mathrm{2}}\:=\frac{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}\:\boldsymbol{\mathrm{cot}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:−\mathrm{tan}\left(\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{B}\right)=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:.\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\frac{\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cotB}+\mathrm{1}}{\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{cotB}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cotB}\:+\mathrm{1}\:=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\left(\mathrm{tan}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}−\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cot}\:\mathrm{B}\right) \\ $$$$\:\:\:\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cotB}\left(\mathrm{1}+\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\right)=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:\mathrm{tan}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}\:−\mathrm{1} \\ $$$$\:\:\:\:\frac{\mathrm{2b}}{\mathrm{b}−\mathrm{c}}\mathrm{cotB}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{cot}\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\therefore\:\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}−\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{cot}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\:=\:\mathrm{2}\boldsymbol{\mathrm{bcotB}}\:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{proved}} \\ $$

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