Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

$$\mathrm{By}\:\mathrm{Napier}'\mathrm{s}\:\mathrm{Analogy}\: \\ $$$$\mathrm{for}\:\mathrm{any}\bigtriangleup\mathrm{ABC} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{B}}−\boldsymbol{\mathrm{C}}}{\mathrm{2}}\:=\frac{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}\:\boldsymbol{\mathrm{cot}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:−\mathrm{tan}\left(\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{B}\right)=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:.\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\frac{\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cotB}+\mathrm{1}}{\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{cotB}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cotB}\:+\mathrm{1}\:=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\left(\mathrm{tan}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}−\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cot}\:\mathrm{B}\right) \\ $$$$\:\:\:\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cotB}\left(\mathrm{1}+\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\right)=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:\mathrm{tan}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}\:−\mathrm{1} \\ $$$$\:\:\:\:\frac{\mathrm{2b}}{\mathrm{b}−\mathrm{c}}\mathrm{cotB}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}−\mathrm{c}}\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{cot}\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\therefore\:\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}−\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{cot}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\:=\:\mathrm{2}\boldsymbol{\mathrm{bcotB}}\:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{proved}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com