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Author: Tinku Tara

Question-73828

Question Number 73828 by ajfour last updated on 16/Nov/19 Commented by ajfour last updated on 17/Nov/19 $${If}\:{on}\:{each}\:{face}\:{of}\:{the}\:{larger}\:{cube} \\ $$$${there}\:{is}\:\left({at}\:{least}\right)\:{one}\:{corner}\:{of}\: \\ $$$${the}\:{inner}\:{cube},\:{then} \\ $$$${find}\:{minimum}\:{value}\:{of}\:{ratio}\:{r}. \\ $$$$\:{r}=\frac{{s}}{{a}}\:=\:\frac{{edge}\:{length}\:{of}\:{inner}\:{cube}}{{edge}\:{length}\:{of}\:{outer}\:{cube}}\:.…

Show-that-tan-tan-tan-1-tan-tan-

Question Number 8287 by lepan last updated on 06/Oct/16 $${Show}\:{that}\:{tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}. \\ $$ Answered by ridwan balatif last updated on 06/Oct/16 $$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{sin}\left(\alpha+\beta\right)}{\mathrm{cos}\left(\alpha+\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha}{\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}{\left(\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}…

n-0-1-6n-

Question Number 139359 by Dwaipayan Shikari last updated on 26/Apr/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}\right)!} \\ $$ Commented by Dwaipayan Shikari last updated on 28/Apr/21 $$\frac{{cosh}\left(\mathrm{1}\right)+\mathrm{2}{cosh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{3}} \\…