Question Number 149184 by liberty last updated on 03/Aug/21 $$\:\:\:\Omega\:=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{tan}\:\sqrt{\mathrm{3x}+\mathrm{2}}−\mathrm{tan}\:\sqrt{\mathrm{2x}+\mathrm{3}}}{\mathrm{tan}\:\sqrt{\mathrm{5x}+\mathrm{4}}−\mathrm{tan}\:\sqrt{\mathrm{4x}+\mathrm{5}}}\:=? \\ $$ Answered by EDWIN88 last updated on 03/Aug/21 $$\Omega=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{3}{x}+\mathrm{2}}}\:\mathrm{sec}\:^{\mathrm{2}} \:\sqrt{\mathrm{3}{x}+\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{3}}}\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{2}{x}+\mathrm{3}}}{\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5}{x}+\mathrm{4}}}\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{5}{x}+\mathrm{4}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{x}+\mathrm{5}}}\mathrm{sec}\:^{\mathrm{2}}…
Question Number 18114 by ibraheem160 last updated on 15/Jul/17 $$\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}{\left.\mathrm{1}−\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}} \\ $$$$\Rightarrow\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta}} \\ $$$$\mathrm{recall}\:\mathrm{that}:\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{1}=\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\theta}+\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} \\ $$$$=\mathrm{cosec}\theta+\mathrm{cot}\theta…
Question Number 149187 by mathdanisur last updated on 03/Aug/21 Commented by mr W last updated on 03/Aug/21 $$=\mathrm{6} \\ $$$${see}\:{Q}\mathrm{74970} \\ $$ Commented by mathdanisur…
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Question Number 83649 by 698148290 last updated on 04/Mar/20 Answered by MJS last updated on 05/Mar/20 $$=\int\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}+\int\frac{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{{x}}{dx}+\int\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{4}} }}{{x}}{dx} \\ $$$$\mathrm{substitute} \\ $$$${u}=\sqrt{\mathrm{1}−{x}}\:\rightarrow\:{dx}=−\mathrm{2}\sqrt{\mathrm{1}−{x}}{du} \\ $$$${v}=\sqrt{\mathrm{2}−{x}^{\mathrm{2}}…
Question Number 18111 by ajfour last updated on 15/Jul/17 Commented by ajfour last updated on 15/Jul/17 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{R}.\:\mathrm{A}\:\mathrm{chord} \\ $$$$\mathrm{AB}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{through}\:\mathrm{point}\:\mathrm{M}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{diameter}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\boldsymbol{\phi}\:\mathrm{to}\:\mathrm{it}; \\ $$$$\:\mathrm{BM}:\mathrm{AM}=\mathrm{p}:\mathrm{q}\:.\:\mathrm{Through}\:\mathrm{point}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{dropped}\:\:\mathrm{a}\:\mathrm{perpendicular}\:\mathrm{BC}\:\mathrm{to}\:\mathrm{the}…
Question Number 149180 by mathdanisur last updated on 03/Aug/21 Answered by Kamel last updated on 03/Aug/21 $${L}=\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{k}}}\right)} =\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}}…
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Question Number 18109 by virus last updated on 15/Jul/17 Commented by virus last updated on 15/Jul/17 $${in}\:{air}\:{no}\:{collision} \\ $$ Commented by virus last updated on…
Question Number 83642 by niroj last updated on 04/Mar/20 $$ \\ $$$$\: \\ $$$$\mathfrak{Find}\:\mathfrak{the}\:\mathfrak{surface}\:\mathfrak{area}\:\mathfrak{of}\:\mathfrak{the}\:\mathfrak{solid}\:\mathfrak{generated} \\ $$$$\:\:\mathfrak{by}\:\mathfrak{the}\:\mathfrak{revolution}\:\mathfrak{of}\:\mathfrak{the}\:\mathfrak{cardioids}\:\mathfrak{r}=\mathfrak{a}\left(\mathrm{1}+\mathfrak{cos}\:\theta\right)\:\mathfrak{about}\:\mathfrak{the}\:\mathfrak{initial}\:\mathfrak{line}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com