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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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