Question Number 143414 by Snail last updated on 14/Jun/21 $${Evaluate}\: \\ $$$$\lfloor\frac{\left\{\underset{{i}=\mathrm{1}} {\overset{\mathrm{1010}} {\sum}}\:{tan}^{\mathrm{2}} \left(\frac{{i}\pi}{\mathrm{2021}}\right)\right\}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\underset{{i}=\mathrm{1}} {\overset{\mathrm{1010}} {\prod}}\:{tan}\:\left(\frac{{i}\pi}{\mathrm{2021}}\right)}\rfloor\:\:\:\:{wherw}\lfloor\centerdot\rfloor\:{denotes}\:{GIF} \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 143410 by mathdanisur last updated on 14/Jun/21 $${if}\:\:{x};{y};{z}>\mathrm{0}\:\:{prove}\:{that}… \\ $$$$\left(\frac{{x}}{{z}}\right)^{\mathrm{2}} \boldsymbol{{e}}^{\left(\frac{{z}}{{x}}\right)^{\mathrm{2}} } +\:\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \boldsymbol{{e}}^{\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} } +\:\left(\frac{{z}}{{y}}\right)^{\mathrm{2}} \:\boldsymbol{{e}}^{\left(\frac{{y}}{{z}}\right)^{\mathrm{2}} } \geqslant\:\mathrm{3}\boldsymbol{{e}} \\ $$ Commented by…
Question Number 77864 by Pratah last updated on 11/Jan/20 Commented by Pratah last updated on 11/Jan/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{inequality} \\ $$ Answered by key of knowledge last…
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Question Number 143384 by mathdanisur last updated on 13/Jun/21 $${proof}:\:\:{tg}^{\mathrm{2}} \left(\mathrm{36}°\right)\:\centerdot\:{tg}^{\mathrm{2}} \left(\mathrm{72}°\right)\:=\:\mathrm{5} \\ $$ Answered by Dwaipayan Shikari last updated on 13/Jun/21 $${tan}\mathrm{36}°=\frac{{sin}\left(\mathrm{36}°\right)}{{cos}\left(\mathrm{36}°\right)}=\frac{\mathrm{1}−{cos}\mathrm{72}°}{{sin}\mathrm{72}°}=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}{\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}=\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}} \\ $$$${tan}\mathrm{72}°=\frac{{cos}\mathrm{18}°}{{sin}\mathrm{18}°}=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}…
Question Number 143368 by mr W last updated on 13/Jun/21 Commented by mr W last updated on 13/Jun/21 $${This}\:{is}\:{a}\:{solved}\:{old}\:{question}\: \\ $$$$\left({Q}\mathrm{142207}\right).\:{The}\:{answer}\:{is} \\ $$$${a}_{{n}} =\frac{\sqrt{\mathrm{2}}\left[\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} +\mathrm{1}\right]}{\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}…
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Question Number 12292 by sin (x) last updated on 18/Apr/17 Answered by mrW1 last updated on 18/Apr/17 $${S}=\frac{\mathrm{1}}{\mathrm{6}}×\pi{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{3}}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 77819 by aliesam last updated on 10/Jan/20 Answered by lémùst last updated on 10/Jan/20 $${posons}\:{t}=\:^{\mathrm{3}} \sqrt{{r}}\:,\:{ainsi}\:{t}\:−\:\frac{\mathrm{1}}{{t}}\:=\:\mathrm{1} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\pm\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${donc}\:{r}=\pm\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}}…
Question Number 77804 by BK last updated on 10/Jan/20 Commented by MJS last updated on 10/Jan/20 $$\mathrm{is}\:\mathrm{2}^{\mid{x}\mid} \:\mathrm{positive}\:\mathrm{or}\:\mathrm{negative}? \\ $$$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{sin}\:{x}^{\mathrm{2}} \:? \\ $$$$\mathrm{what}\:\mathrm{are}\:{your}\:\mathrm{thoughts},\:\mathrm{Sir}\:\mathrm{BK}? \\ $$…