Question Number 108573 by mnjuly1970 last updated on 17/Aug/20
$$\:\:\:\:\:\:\:\:\:\mathrm{please}:\:\:\:\mathrm{in}\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\:\frac{{cos}\left(\mathrm{A}\right)}{{sin}\left(\mathrm{B}\right){sin}\left(\mathrm{C}\right)}\:+\frac{{cos}\left(\mathrm{B}\right)}{{sin}\left(\mathrm{A}\right){sin}\left(\mathrm{C}\right)}+\frac{{cos}\left(\mathrm{C}\right)}{{sin}\left(\mathrm{A}\right){sin}\left(\mathrm{B}\right)}\:=\mathrm{2}\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by veth last updated on 17/Aug/20
Commented by mnjuly1970 last updated on 17/Aug/20
$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{master}.\mathrm{mercey}... \\ $$
Answered by ajfour last updated on 17/Aug/20
![to prove Σ((sin Acos A)/(sin Asin Bsin C))=2 l.h.s.=(1/2)(((Σsin 2A)/(Πsin A))) =(1/2)(((2sin (A+B)cos (A−B)+sin 2C)/(Πsin A))) =((cos (A−B)+cos C)/(sin Asin B)) =2[((cos (A−B)+cos C)/(cos (A−B)−cos (A+B)))] =2[((cos (A−B)+cos C)/(cos (A−B)+cos C))] = 2 .](Q108580.png)
$${to}\:{prove}\:\:\:\:\:\Sigma\frac{\mathrm{sin}\:{A}\mathrm{cos}\:{A}}{\mathrm{sin}\:{A}\mathrm{sin}\:{B}\mathrm{sin}\:{C}}=\mathrm{2} \\ $$$${l}.{h}.{s}.=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Sigma\mathrm{sin}\:\mathrm{2}{A}}{\Pi\mathrm{sin}\:{A}}\right) \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{sin}\:\mathrm{2}{C}}{\Pi\mathrm{sin}\:{A}}\right) \\ $$$$=\frac{\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{cos}\:{C}}{\mathrm{sin}\:{A}\mathrm{sin}\:{B}} \\ $$$$=\mathrm{2}\left[\frac{\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{cos}\:{C}}{\mathrm{cos}\:\left({A}−{B}\right)−\mathrm{cos}\:\left({A}+{B}\right)}\right] \\ $$$$=\mathrm{2}\left[\frac{\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{cos}\:{C}}{\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{cos}\:{C}}\right]\:\:=\:\mathrm{2}\:. \\ $$
Commented by mnjuly1970 last updated on 17/Aug/20
$$\mathrm{god}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{master}\:.\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\ $$
Answered by nimnim last updated on 17/Aug/20
![A+B+C=π. LHS((cosA)/(sinBsinC))−((cos(B+C))/(sinAsinC))−((cos(A+B))/(sinAsinB)) =((cosA)/(sinBsinC))−[((cosAcosC−sinAsinC)/(sinAsinC))+((cosAcosB−sinAsinB)/(sinAsinB))] =((cosA)/(sinBsinC))−[((cosAcosC)/(sinAsinC))−1+((cosAcosB)/(sinAsinB))−1] =((cosA)/(sinBsinC))−cosA[((sinBcosC+cosBsinC)/(sinAsinBsinC))]+2 =((cosA)/(sinBsinC))−((cosAsin(B+C))/(sin(B+C)sinBsinC))+2 =((cosA)/(sinBsinC))−((cosA)/(sinBsinC))+2 =2(RHS)](Q108585.png)
$$\:\:\:\:{A}+{B}+{C}=\pi.\: \\ $$$${LHS}\frac{{cosA}}{{sinBsinC}}−\frac{{cos}\left({B}+{C}\right)}{{sinAsinC}}−\frac{{cos}\left({A}+{B}\right)}{{sinAsinB}} \\ $$$$\:\:=\frac{{cosA}}{{sinBsinC}}−\left[\frac{{cosAcosC}−{sinAsinC}}{{sinAsinC}}+\frac{{cosAcosB}−{sinAsinB}}{{sinAsinB}}\right] \\ $$$$\:\:=\frac{{cosA}}{{sinBsinC}}−\left[\frac{{cosAcosC}}{{sinAsinC}}−\mathrm{1}+\frac{{cosAcosB}}{{sinAsinB}}−\mathrm{1}\right] \\ $$$$\:\:=\frac{{cosA}}{{sinBsinC}}−{cosA}\left[\frac{{sinBcosC}+{cosBsinC}}{{sinAsinBsinC}}\right]+\mathrm{2} \\ $$$$\:\:=\frac{{cosA}}{{sinBsinC}}−\frac{{cosAsin}\left({B}+{C}\right)}{{sin}\left({B}+{C}\right){sinBsinC}}+\mathrm{2} \\ $$$$\:\:=\frac{{cosA}}{{sinBsinC}}−\frac{{cosA}}{{sinBsinC}}+\mathrm{2} \\ $$$$\:\:=\mathrm{2}\left({RHS}\right) \\ $$
Commented by mnjuly1970 last updated on 17/Aug/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear} \\ $$$$\mathrm{friend}.\mathrm{peace}\:\mathrm{be}\:\mathrm{upon}\:\mathrm{you}\:... \\ $$