Question Number 99742 by Ar Brandon last updated on 23/Jun/20 $$\mathcal{G}\mathrm{iven}\:\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}°\:\:\mathrm{prove}\:\:\mathrm{that} \\ $$$$\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{1} \\ $$ Answered by smridha last updated on 23/Jun/20 $$\boldsymbol{{tan}}\left(\frac{\boldsymbol{{A}}}{\mathrm{2}}+\frac{\boldsymbol{{B}}}{\mathrm{2}}+\frac{\boldsymbol{{C}}}{\mathrm{2}}\right)=\boldsymbol{{tan}}\frac{\boldsymbol{\pi}}{\mathrm{2}}=\infty\: \\ $$$$\Rightarrow\frac{\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}−\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}}{\mathrm{1}−\left[\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}\right]}=\infty…
Question Number 165270 by cortano1 last updated on 28/Jan/22 Answered by TheSupreme last updated on 28/Jan/22 $${B}+{C}=\mathrm{120} \\ $$$$\frac{{sin}\left({C}\right)}{{sin}\left({B}\right)}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${sin}\left({C}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){sin}\left(\mathrm{120}−{C}\right) \\ $$$${sin}\left({C}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\frac{{cos}\left({C}\right)}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\left({C}\right)\right) \\ $$$${sin}\left({C}\right)\left(\mathrm{1}+\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\frac{{cos}\left({C}\right)}{\mathrm{2}}…
Question Number 99680 by Algoritm last updated on 22/Jun/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 34117 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18 Commented by abdo imad last updated on 01/May/18 $${let}\:{put}\:{f}\left({x}\right)=\frac{{sin}\theta}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:\:{poles}\:{of}\:{f}\left({x}\right)? \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}=\mathrm{0}\: \\ $$$$\Delta^{'} ={cos}^{\mathrm{2}}…
Question Number 34114 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 34115 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18 Commented by abdo mathsup 649 cc last updated on 03/May/18 $${we}\:{have}\:\:\:{forf}\left({t}\right)={arctant} \\ $$$${f}^{'} \left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:{with}\:−\mathrm{1}<{t}<\mathrm{1}\Rightarrow{f}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}}…
Question Number 34116 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18 Commented by abdo mathsup 649 cc last updated on 02/May/18 $${let}\:{put}\:{f}\left({x}\right)=\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:{the}\:{poles}\:{of}\:{f}\:{are} \\ $$$${x}_{\mathrm{1}} =\:\frac{\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 34113 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 99620 by Dwaipayan Shikari last updated on 22/Jun/20 $${If}\:\:\alpha=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:{then}\:{what}\:{is}\:{the}\:{value}\:{of}\:\left({sin}\alpha{sin}\mathrm{2}\alpha{sin}\mathrm{4}\alpha\right) \\ $$ Commented by Dwaipayan Shikari last updated on 22/Jun/20 $${I}\:{have}\:{found}\:{a}\:{way}\:{to}\:{solve}\:{it} \\ $$$${Suppose} \\…
Question Number 34066 by 5a3k last updated on 30/Apr/18 $$\mathrm{show}\:\mathrm{that}− \\ $$$$\mathrm{sin}\:\mathrm{10}−\sqrt{\mathrm{3}}\mathrm{sec10}=\mathrm{4}. \\ $$ Answered by math1967 last updated on 30/Apr/18 $${correct}\:{q}.\:{is}\:\frac{\mathrm{1}}{{sin}\mathrm{10}°}\:−\sqrt{\mathrm{3}\:\:\:}\:{sec}\mathrm{10}° \\ $$$$\frac{{cos}\mathrm{10}°−\sqrt{\mathrm{3}}\:{sin}\mathrm{10}}{{sin}\mathrm{10}{cos}\mathrm{10}\:\:} \\…