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Question Number 16358 by Tinkutara last updated on 21/Jun/17
In any triangle ABC, a cot A + b cot B + c cot C is equal to (1) r + R (2) r − R (3) 2(r + R) (4) 2(r − R)
$$\mathrm{In}\:\mathrm{any}\:\mathrm{triangle}\:{ABC},\:{a}\:\mathrm{cot}\:{A}\:+\:{b}\:\mathrm{cot}\:{B} \\ $$$$+\:{c}\:\mathrm{cot}\:{C}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{1}\right)\:{r}\:+\:{R} \\ $$$$\left(\mathrm{2}\right)\:{r}\:−\:{R} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}\left({r}\:+\:{R}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\left({r}\:−\:{R}\right) \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17
answer (3).
$${answer}\:\left(\mathrm{3}\right). \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17
acotgA=((2RsinA.cosA)/(sinA))=2RcosA ⇒LHS=2R(cosA+cosB+cosC) ΣcosA=2cos((A+B)/2).cos((A−B)/2)+cos(180−A−B)= =2cos((A+B)/2).cos((A−B)/2)−2cos^2 ((A+B)/2)+1= 2cos((A+B)/2)(cos((A−B)/2)−cos((A+B)/2))+1= =2cos((180−C)/2).2sin(A/2)sin(B/2)+1= =4sin(A/2)sin(B/2)sin(C/2)+1 sin(A/2)=(√((1−cosA)/2))=(√((1−((b^2 +c^2 −a^2 )/(2bc)))/2))= =(√(((a+b−c)(a+c−b))/(4bc)))=(√(((p−b)(p−c))/(bc))) ⇒Πsin(A/2)=(√(((p−b)(p−c)(p−a)(p−c)(p−a)(p−b))/(a^2 b^2 c^2 )))= =(((p−a)(p−b)(p−c))/(abc))=((S^2 /p)/(4R.S))=(S/(4R.p))=(r/(4R)) ⇒LHS=2R(4×(r/(4R))+1)=2r+2R=2(r+R) .■
$${acotgA}=\frac{\mathrm{2}{RsinA}.{cosA}}{{sinA}}=\mathrm{2}{RcosA} \\ $$$$\Rightarrow{LHS}=\mathrm{2}{R}\left({cosA}+{cosB}+{cosC}\right) \\ $$$$\Sigma{cosA}=\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}.{cos}\frac{{A}−{B}}{\mathrm{2}}+{cos}\left(\mathrm{180}−{A}−{B}\right)= \\ $$$$=\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}.{cos}\frac{{A}−{B}}{\mathrm{2}}−\mathrm{2}{cos}^{\mathrm{2}} \frac{{A}+{B}}{\mathrm{2}}+\mathrm{1}= \\ $$$$\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}\left({cos}\frac{{A}−{B}}{\mathrm{2}}−{cos}\frac{{A}+{B}}{\mathrm{2}}\right)+\mathrm{1}= \\ $$$$=\mathrm{2}{cos}\frac{\mathrm{180}−{C}}{\mathrm{2}}.\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}+\mathrm{1}= \\ $$$$=\mathrm{4}{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}+\mathrm{1} \\ $$$${sin}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−{cosA}}{\mathrm{2}}}=\sqrt{\frac{\mathrm{1}−\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{\mathrm{2}}}= \\ $$$$=\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)}{\mathrm{4}{bc}}}=\sqrt{\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{bc}}} \\ $$$$\Rightarrow\Pi{sin}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\left({p}−{b}\right)\left({p}−{c}\right)\left({p}−{a}\right)\left({p}−{c}\right)\left({p}−{a}\right)\left({p}−{b}\right)}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }}= \\ $$$$=\frac{\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)}{{abc}}=\frac{\frac{{S}^{\mathrm{2}} }{{p}}}{\mathrm{4}{R}.{S}}=\frac{{S}}{\mathrm{4}{R}.{p}}=\frac{{r}}{\mathrm{4}{R}} \\ $$$$\Rightarrow{LHS}=\mathrm{2}{R}\left(\mathrm{4}×\frac{{r}}{\mathrm{4}{R}}+\mathrm{1}\right)=\mathrm{2}{r}+\mathrm{2}{R}=\mathrm{2}\left({r}+{R}\right)\:.\blacksquare \\ $$
Commented by Tinkutara last updated on 22/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17
note: sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2) cos(A/2)=(√((1+cosA)/2))=(√((1+((b^2 +c^2 −a^2 )/(2bc)))/2))= =(√(((b+c+a)(b+c−a))/(4bc)))=(√((p(p−a))/(bc))) ⇒Πcos(A/2)=(√((p(p−a)p(p−b)p(p−c))/(a^2 b^2 c^2 )))= =((p.S)/(abc))=((p.S)/(4R.S))=(p/(4R)) tgA+tgB+tg(180−^((=C)) A−B)= =tgA+tgB−((tgA+tgB)/(1−tgAtgB))=(tgA+tgB)(1−(1/(1−tgA.tgB)))= =−tgA.tgB.((tgA+tgB)/(1−tgA.tgB))=−tgA.tgB.(−tgC)= =tgA.tgB.tgC ΣsinA=4Πcos(A/2) ΣcosA=1+4Πsin(A/2) ΣtgA=ΠtgA
$${note}: \\ $$$${sinA}+{sinB}+{sinC}=\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}} \\ $$$${cos}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}+{cosA}}{\mathrm{2}}}=\sqrt{\frac{\mathrm{1}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{\mathrm{2}}}= \\ $$$$=\sqrt{\frac{\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right)}{\mathrm{4}{bc}}}=\sqrt{\frac{{p}\left({p}−{a}\right)}{{bc}}} \\ $$$$\Rightarrow\Pi{cos}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{{p}\left({p}−{a}\right){p}\left({p}−{b}\right){p}\left({p}−{c}\right)}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }}= \\ $$$$=\frac{{p}.{S}}{{abc}}=\frac{{p}.{S}}{\mathrm{4}{R}.{S}}=\frac{{p}}{\mathrm{4}{R}} \\ $$$${tgA}+{tgB}+{tg}\left(\mathrm{180}\overset{\left(={C}\right)} {−}{A}−{B}\right)= \\ $$$$={tgA}+{tgB}−\frac{{tgA}+{tgB}}{\mathrm{1}−{tgAtgB}}=\left({tgA}+{tgB}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{tgA}.{tgB}}\right)= \\ $$$$=−{tgA}.{tgB}.\frac{{tgA}+{tgB}}{\mathrm{1}−{tgA}.{tgB}}=−{tgA}.{tgB}.\left(−{tgC}\right)= \\ $$$$={tgA}.{tgB}.{tgC} \\ $$$$\Sigma{sinA}=\mathrm{4}\Pi{cos}\frac{{A}}{\mathrm{2}} \\ $$$$\Sigma{cosA}=\mathrm{1}+\mathrm{4}\Pi{sin}\frac{{A}}{\mathrm{2}} \\ $$$$\Sigma{tgA}=\Pi{tgA} \\ $$$$ \\ $$

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