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 $${correct}\:{q}.\:{is}\:\frac{\mathrm{1}}{{sin}\mathrm{10}°}\:−\sqrt{\mathrm{3}\:\:\:}\:{sec}\mathrm{10}° \\ $$$$\frac{{cos}\mathrm{10}°−\sqrt{\mathrm{3}}\:{sin}\mathrm{10}}{{sin}\mathrm{10}{cos}\mathrm{10}\:\:} \\…
Question Number 34064 by math1967 last updated on 30/Apr/18 Answered by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18 $${let}\:{sin}^{\mathrm{2}} \alpha={t} \\ $$$$\frac{{t}^{\mathrm{2}} }{{a}}+\frac{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{{b}}=\frac{\mathrm{1}}{{a}+{b}} \\ $$$${bt}^{\mathrm{2}} +{a}\left(\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}}…
Question Number 99539 by Ar Brandon last updated on 21/Jun/20 $$\mathrm{If}\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}°\:\mathrm{Prove} \\ $$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=\mathrm{4cos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Acos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Bcos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{C} \\ $$ Answered by Dwaipayan Shikari last updated on 22/Jun/20 $${sinA}+{sinB}+{sinC}=\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{B}+{C}}{\mathrm{2}}{cos}\frac{{B}−{C}}{\mathrm{2}} \\…