Question Number 43363 by rahul 19 last updated on 10/Sep/18 $$\mathrm{If}\:\mathrm{z}=\:\mathrm{cos}\:\theta\:+\:\mathrm{isin}\:\theta\:,\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{6}}\:,\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{argument}\:\mathrm{of}\:\:\mathrm{1}−\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}\:. \\ $$ Commented by maxmathsup by imad last updated on 10/Sep/18…
Question Number 174424 by mnjuly1970 last updated on 31/Jul/22 $$ \\ $$$$\:\:\:\:#\:\:{prove}\:\:{that}\:\:# \\ $$$$\:\:\:{tan}^{\:\mathrm{2}} \left(\:\mathrm{10}^{\:\mathrm{o}} \right)\:+\:{tan}^{\:\mathrm{2}} \left(\mathrm{50}^{\:\mathrm{o}} \:\right)+{tan}^{\:\mathrm{2}} \left(\mathrm{70}^{\:\mathrm{o}} \right)\overset{?} {=}\:\mathrm{9} \\ $$$$\:\:\:\:\:−−−−− \\ $$$$…
Question Number 174427 by mnjuly1970 last updated on 31/Jul/22 $$ \\ $$$$\:\:\:{solve}\:\:{for}\:\:\:{x}\:\in\:\mathbb{R}\:: \\ $$$$ \\ $$$$\:\sqrt[{\mathrm{4}}]{\mathrm{12}+\mathrm{4}{x}^{\:} −{x}^{\:\mathrm{2}} }\:+\sqrt{\mathrm{1}+\mathrm{8}{x}−\mathrm{2}{x}^{\:\mathrm{2}} }\:=\:\mathrm{2}{x}^{\:\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{13}\:\:\blacksquare \\ $$$$ \\ $$$$ \\…
Question Number 43353 by pieroo last updated on 10/Sep/18 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{f}\bullet\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}, \\ $$$$\mathrm{find}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$ Commented by maxmathsup by imad last updated on 10/Sep/18 $${we}\:{have}\:{f}\left({x}\right)={y}\:\Rightarrow\mathrm{2}{x}−\mathrm{1}={y}\:\Rightarrow{x}\:=\frac{{y}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}^{−\mathrm{1}}…
Question Number 43349 by peter frank last updated on 10/Sep/18 Answered by math1967 last updated on 10/Sep/18 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{5}}+\sqrt{\left(\mathrm{3}−\sqrt{\left.\mathrm{5}\right)^{\mathrm{2}} }\right.}}} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{3}−\sqrt{\mathrm{5}}}} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{5}}−\sqrt{\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)^{\mathrm{2}} }} \\…
Question Number 43341 by pieroo last updated on 10/Sep/18 $$\mathrm{write}\:\mathrm{2}×\mathrm{7}+\mathrm{3}×\mathrm{8}\:\mathrm{4}×\mathrm{9}+\mathrm{5}×\mathrm{10}+\mathrm{6}×\mathrm{11}\:\mathrm{using}\:\mathrm{the}\:\mathrm{sigma} \\ $$$$\mathrm{notation} \\ $$ Answered by $@ty@m last updated on 10/Sep/18 $$\underset{{n}=\mathrm{2}} {\overset{\mathrm{6}} {\sum}}{n}\left({n}+\mathrm{5}\right) \\…
Question Number 174408 by BaliramKumar last updated on 31/Jul/22 $$\frac{\mathrm{2525}}{{x}}\:\equiv\:\mathrm{5}\:\left({remainder}\right),\:{How}\:{many}\:{value}\:{of}\:\boldsymbol{{x}}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108870 by 150505R last updated on 19/Aug/20 Commented by Dwaipayan Shikari last updated on 19/Aug/20 $${cot}\frac{\pi}{\mathrm{2}{n}} \\ $$ Answered by Ar Brandon last…
Question Number 108871 by 150505R last updated on 19/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 174377 by AgniMath last updated on 31/Jul/22 Answered by okbruh123 last updated on 31/Jul/22 $$\mathrm{ez}\:\mathrm{telescoping}\:\mathrm{lol} \\ $$$$\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{5}}+…+\frac{\mathrm{1}}{\mathrm{11}\centerdot\mathrm{13}} \\ $$$$\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{5}−\mathrm{3}}{\mathrm{3}\centerdot\mathrm{5}}+…+\frac{\mathrm{13}−\mathrm{11}}{\mathrm{11}\centerdot\mathrm{13}}\right) \\ $$$$\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+…+\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{1}}{\mathrm{13}}\right) \\ $$$$\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{13}}\right)=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{10}}{\mathrm{39}}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{39}\:}\:\mathrm{now}\:\mathrm{do}\:\mathrm{basic}\:\mathrm{maths}…