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Given-A-B-C-180-prove-that-tan-A-2-tan-B-2-tan-B-2-tan-C-2-tan-C-2-tan-A-2-1-




Question Number 99742 by Ar Brandon last updated on 23/Jun/20
Given  A+B+C=180°  prove  that  tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1
$$\mathcal{G}\mathrm{iven}\:\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}°\:\:\mathrm{prove}\:\:\mathrm{that} \\ $$$$\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{1} \\ $$
Answered by smridha last updated on 23/Jun/20
tan((A/2)+(B/2)+(C/2))=tan(𝛑/2)=∞   ⇒((tan(A/2)+tan(B/2)+tan(C/2)−tan(A/2).tan(B/2)tan(C/2))/(1−[tan(A/2).tan(B/2)+tan(B/2).tan(C/2)+tan(C/2).tan(A/2)]))=∞  this is possible when the denominator  goes to   0  therefore..  [tan(A/2)tan(B/2)+tan(B/2).tan(C/2)+tan(C/2).tan(A/2)]=1
$$\boldsymbol{{tan}}\left(\frac{\boldsymbol{{A}}}{\mathrm{2}}+\frac{\boldsymbol{{B}}}{\mathrm{2}}+\frac{\boldsymbol{{C}}}{\mathrm{2}}\right)=\boldsymbol{{tan}}\frac{\boldsymbol{\pi}}{\mathrm{2}}=\infty\: \\ $$$$\Rightarrow\frac{\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}−\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}}{\mathrm{1}−\left[\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}\right]}=\infty \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{possible}}\:\boldsymbol{{when}}\:\boldsymbol{{the}}\:\boldsymbol{{denominator}} \\ $$$$\boldsymbol{{goes}}\:\boldsymbol{{to}}\:\:\:\mathrm{0} \\ $$$$\boldsymbol{{therefore}}.. \\ $$$$\left[\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}\right]=\mathrm{1} \\ $$$$ \\ $$
Commented by smridha last updated on 23/Jun/20
is there any logic behind your  silly question??
$$\boldsymbol{{is}}\:\boldsymbol{{there}}\:\boldsymbol{{any}}\:\boldsymbol{{logic}}\:\boldsymbol{{behind}}\:\boldsymbol{{your}} \\ $$$$\boldsymbol{{silly}}\:\boldsymbol{{question}}?? \\ $$
Commented by Ar Brandon last updated on 23/Jun/20
Thank you Sir
Commented by smridha last updated on 23/Jun/20
yeah!! welcome
Commented by Rasheed.Sindhi last updated on 23/Jun/20
Sir, is there any logic behind   writing in different colours?
$${Sir},\:{is}\:{there}\:{any}\:{logic}\:{behind}\: \\ $$$${writing}\:{in}\:{different}\:{colours}? \\ $$
Commented by Rasheed.Sindhi last updated on 23/Jun/20
If  you do something different,  you should have a reason for that.  Specially if you are math-related   person.May be only beatification  be a reason.
$${If}\:\:{you}\:{do}\:{something}\:{different}, \\ $$$${you}\:{should}\:{have}\:{a}\:{reason}\:{for}\:{that}. \\ $$$${Specially}\:{if}\:{you}\:{are}\:{math}-{related}\: \\ $$$${person}.{May}\:{be}\:\boldsymbol{{only}}\:\boldsymbol{{beatification}} \\ $$$${be}\:{a}\:{reason}. \\ $$
Commented by smridha last updated on 23/Jun/20
why obsessed for black and white  man!!there is a facility for  choosing defrent colours ..you may  use or you may not..that′s your  choice...which colour combination  I used that′s my business...  the solution is prominent and  bolt enough so there shall be no  objection....
$$\boldsymbol{{why}}\:\boldsymbol{{obsessed}}\:\boldsymbol{{for}}\:\boldsymbol{{black}}\:\boldsymbol{{and}}\:\boldsymbol{{white}} \\ $$$$\boldsymbol{{man}}!!\boldsymbol{{there}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{facility}}\:\boldsymbol{{for}} \\ $$$$\boldsymbol{{choosing}}\:\boldsymbol{{defrent}}\:\boldsymbol{{colours}}\:..\boldsymbol{{you}}\:\boldsymbol{{may}} \\ $$$$\boldsymbol{{use}}\:\boldsymbol{{or}}\:\boldsymbol{{you}}\:\boldsymbol{{may}}\:\boldsymbol{{not}}..\boldsymbol{{that}}'{s}\:\boldsymbol{{your}} \\ $$$$\boldsymbol{{choice}}…\boldsymbol{{which}}\:\boldsymbol{{colour}}\:\boldsymbol{{combination}} \\ $$$${I}\:\boldsymbol{{used}}\:\boldsymbol{{that}}'{s}\:\boldsymbol{{my}}\:\boldsymbol{{business}}… \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{solution}}\:\boldsymbol{{is}}\:\boldsymbol{{prominent}}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{bolt}}\:\boldsymbol{{enough}}\:\boldsymbol{{so}}\:\boldsymbol{{there}}\:\boldsymbol{{shall}}\:\boldsymbol{{be}}\:\boldsymbol{{no}} \\ $$$$\boldsymbol{{objection}}…. \\ $$
Commented by Rasheed.Sindhi last updated on 23/Jun/20
Ok sIR aS Y⊚_U  WiS^h  !  I only want to say that use of  different colors should be helpful  in understanding process.It should  be used as a tool.
$${Ok}\:\boldsymbol{\mathrm{s}}{I}\mathbb{R}\:\mathrm{a}\mathcal{S}\:\mathcal{Y}\circledcirc_{\boldsymbol{\mathrm{U}}} \:\mathcal{W}{i}\mathcal{S}^{{h}} \:! \\ $$$${I}\:{only}\:{want}\:{to}\:{say}\:{that}\:{use}\:{of} \\ $$$${different}\:{colors}\:{should}\:{be}\:{helpful} \\ $$$${in}\:{understanding}\:{process}.{It}\:{should} \\ $$$${be}\:{used}\:{as}\:{a}\:{tool}. \\ $$
Commented by Ar Brandon last updated on 23/Jun/20
��cheers
Commented by Rasheed.Sindhi last updated on 24/Jun/20
��cool !
Commented by Rasheed.Sindhi last updated on 24/Jun/20
Sir/Madam Smridha  Sorry if my coool words made you  angry.Actually I didn′t want that.
$${Sir}/{Madam}\:{Smridha} \\ $$$${Sorry}\:{if}\:{my}\:{coool}\:{words}\:{made}\:{you} \\ $$$$\boldsymbol{{angry}}.{Actually}\:{I}\:{didn}'{t}\:{want}\:{that}. \\ $$
Commented by smridha last updated on 24/Jun/20
this is not for cool words   this is for your previous comment.  this will be good if you focus  on solving problem instead of  creating violence.I donot know  how long you are present in this  forum...but how much I know  there is no madam!!that′s your  VERDENCY...
$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\boldsymbol{{for}}\:\boldsymbol{{cool}}\:\boldsymbol{{words}}\: \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{for}}\:\boldsymbol{{your}}\:\boldsymbol{{previous}}\:\boldsymbol{{comment}}. \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{will}}\:\boldsymbol{{be}}\:\boldsymbol{{good}}\:\boldsymbol{{if}}\:\boldsymbol{{you}}\:\boldsymbol{{focus}} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{solving}}\:\boldsymbol{{problem}}\:\boldsymbol{{instead}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{creating}}\:\boldsymbol{{violence}}.\boldsymbol{{I}}\:{do}\boldsymbol{{no}\mathrm{t}}\:\boldsymbol{{know}} \\ $$$$\boldsymbol{{how}}\:\boldsymbol{{long}}\:\boldsymbol{{you}}\:\boldsymbol{{are}}\:\boldsymbol{{present}}\:\boldsymbol{{in}}\:\boldsymbol{{this}} \\ $$$$\boldsymbol{{forum}}…\boldsymbol{{but}}\:\boldsymbol{{how}}\:\boldsymbol{{much}}\:\boldsymbol{{I}}\:\boldsymbol{{know}} \\ $$$$\boldsymbol{{there}}\:\boldsymbol{{is}}\:\boldsymbol{{no}}\:\boldsymbol{{madam}}!!\boldsymbol{{that}}'\boldsymbol{{s}}\:\boldsymbol{{your}} \\ $$$$\mathbb{VERDENCY}… \\ $$
Commented by Rasheed.Sindhi last updated on 24/Jun/20
Sir Smridha, pl go and work  smoothly.Write your colorful/  beautiful answers.No one will  disturb you!..When I scroll through  posts of the forum,I imediately  recognise yor posts without reading  your name!....  Actually I wasn′t certain of your  being Mr. so I write both sir &  madam. There was no other   reason.Don′t think negatively  for others.Thank you.Bye!  (The thread has been too long!)
$$\boldsymbol{{Sir}}\:{Smridha},\:{pl}\:{go}\:{and}\:{work} \\ $$$${smoothly}.{Write}\:{your}\:{colorful}/ \\ $$$${beautiful}\:{answers}.{No}\:{one}\:{will} \\ $$$${disturb}\:{you}!..{When}\:{I}\:{scroll}\:{through} \\ $$$${posts}\:{of}\:{the}\:{forum},{I}\:{imediately} \\ $$$${recognise}\:{yor}\:{posts}\:{without}\:{reading} \\ $$$${your}\:{name}!…. \\ $$$${Actually}\:{I}\:{wasn}'{t}\:{certain}\:{of}\:{your} \\ $$$${being}\:{Mr}.\:{so}\:{I}\:{write}\:{both}\:{sir}\:\& \\ $$$${madam}.\:{There}\:{was}\:{no}\:{other}\: \\ $$$${reason}.{Don}'{t}\:{think}\:{negatively} \\ $$$${for}\:{others}.{Thank}\:{you}.{Bye}! \\ $$$$\left({The}\:{thread}\:{has}\:{been}\:{too}\:{long}!\right) \\ $$
Answered by Dwaipayan Shikari last updated on 23/Jun/20
tan(B/2)[((sin(((A+C))/2))/(cos(C/2)cos(A/2)))]+((sin(A/2)sin(C/2))/(cos(A/2)cos(C/2)))   {{{((sin(A/2))/(cos(A/2)))+((sin(C/2))/(cos(C/2)))=((sin(((A+C))/2))/(cos(A/2)cos(C/2)))}}}  =((sin((π/2)−((A+C)/2))+sin(A/2)sin(C/2))/(cos(A/2)cos(C/2)))=((cos(A/2)cos(C/2))/(cos(A/2)cos(C/2)))=1[Proved]{{{Because tan(A/2)tan(B/2)+tan(C/2)tan(B/2)=tan(B/2)(((sin(A+C))/(cos(A/2)cos(C/2))))                                                                                                       {sin((A+C)/2)=cos(B/2)                                                                                                       {sin((π/2)−((A+C)/2))                                                                                          =cos(A/2)cos(C/2)−sin(A/2)sin(C/2)
$${tan}\frac{{B}}{\mathrm{2}}\left[\frac{{sin}\frac{\left({A}+{C}\right)}{\mathrm{2}}}{{cos}\frac{{C}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}}}\right]+\frac{{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}}{{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}}\:\:\:\left\{\left\{\left\{\frac{{sin}\frac{{A}}{\mathrm{2}}}{{cos}\frac{{A}}{\mathrm{2}}}+\frac{{sin}\frac{{C}}{\mathrm{2}}}{{cos}\frac{{C}}{\mathrm{2}}}=\frac{{sin}\frac{\left({A}+{C}\right)}{\mathrm{2}}}{{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}}\right\}\right\}\right\} \\ $$$$=\frac{{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{C}}{\mathrm{2}}\right)+{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}}{{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}}=\frac{{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}}{{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}}=\mathrm{1}\left[{Proved}\right]\left\{\left\{\left\{{Because}\:{tan}\frac{{A}}{\mathrm{2}}{tan}\frac{{B}}{\mathrm{2}}+{tan}\frac{{C}}{\mathrm{2}}{tan}\frac{{B}}{\mathrm{2}}={tan}\frac{{B}}{\mathrm{2}}\left(\frac{{sin}\left({A}+{C}\right)}{{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}}\right)\right.\right.\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{sin}\frac{{A}+{C}}{\mathrm{2}}={cos}\frac{{B}}{\mathrm{2}}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{C}}{\mathrm{2}}\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}−{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Ar Brandon last updated on 23/Jun/20
Thank you Sir
Commented by Dwaipayan Shikari last updated on 23/Jun/20
please don't tell me sir .I am a student. I am in pleasure to solve your problem . Thanking you.
Commented by Ar Brandon last updated on 23/Jun/20
As you wish ! ��
Answered by 1549442205 last updated on 23/Jun/20
We have ((A+B)/2)=(π/2)−(C/2)⇒tan((A+B)/2)=tan((π/2)−(C/2))  ⇒((tan(A/2)+tan(B/2))/(1−tan(A/2)tan(B/2)))=cot(C/2)=(1/(tan(C/2)))   ⇒tan(A/2)tan(C/2)+tan(B/2)tan(C/2)+tan(A/2)tan(B/2)=1(q.e.d)
$$\mathrm{We}\:\mathrm{have}\:\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{C}}{\mathrm{2}}\Rightarrow\mathrm{tan}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\mathrm{tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{C}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}}=\mathrm{cot}\frac{\mathrm{C}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}} \\ $$$$\:\Rightarrow\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}=\mathrm{1}\left(\mathrm{q}.\mathrm{e}.\mathrm{d}\right) \\ $$
Commented by Ar Brandon last updated on 23/Jun/20
wow ! that was brilliant. Thanks
Commented by 1549442205 last updated on 25/Jun/20
you are wellcome,sir!
$$\mathrm{you}\:\mathrm{are}\:\mathrm{wellcome},\mathrm{sir}! \\ $$

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