Question Number 63769 by aliesam last updated on 08/Jul/19 Commented by mathmax by abdo last updated on 09/Jul/19 $${let}\:{prove}\:{by}\:{recurence}\:{n}=\mathrm{0}\:\:\:\:{A}_{\mathrm{0}} =\mathrm{0}\:{is}\:{divisible}\:{by}\:\mathrm{6} \\ $$$${let}\:{suppose}\:{A}_{{n}} ={n}^{\mathrm{3}} \:+\mathrm{5}{n}\:{is}\:{divisible}\:{by}\:\mathrm{6}\:\Rightarrow \\…
Question Number 63763 by aliesam last updated on 08/Jul/19 Commented by Prithwish sen last updated on 08/Jul/19 $$\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:+\:\mathrm{i}\:\mathrm{sin}\left(\mathrm{2x}\right)}\:=\:\frac{\mathrm{2}\left\{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)\right\}}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\:+\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)}{\mathrm{1}+\:\mathrm{cos}\left(\mathrm{2x}\right)}\:=\:\mathrm{1}\:−\:\mathrm{i}\frac{\mathrm{2sinx}.\mathrm{cosx}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{1}\:−\mathrm{itanx}\:\mathrm{Hence}\:\mathrm{proved}.…
Question Number 63751 by Tawa1 last updated on 08/Jul/19 $$\mathrm{Where}\:\mathrm{is}\:\mathrm{sir}\:\mathrm{tanmay} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129271 by MathSh last updated on 14/Jan/21 $${In}\:{the}\:{I}\:{park}\:\mathrm{50\%}\:{of}\:{the}\:{schools} \\ $$$${students},\:\mathrm{80\%}\:{in}\:{the}\:{II}\:{park}, \\ $$$$\mathrm{90\%}\:{has}\:{partic}\imath{pated}\:{in}\:{the}\:{III}\:{park}. \\ $$$$\mathrm{160}\:{students}\:{have}\:{partic}\imath{pated}\:{in} \\ $$$${all}\:\mathrm{3}\:{parks},\:{the}\:{rest}\:{have} \\ $$$${partic}\imath{pated}\:{in}\:\mathrm{2}\:{park}. \\ $$$${How}\:{many}\:{students}\:{are}\:{there}\:{in} \\ $$$${the}\:{school}\:?\: \\…
Question Number 129242 by Adel last updated on 14/Jan/21 $$\mathrm{cot}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}+\mathrm{cot}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{cot}^{\mathrm{2}} \frac{\mathrm{7}\pi}{\mathrm{2}}=? \\ $$$$\mathrm{answer}\:\:\mathrm{by}\:\mathrm{solve} \\ $$ Commented by liberty last updated on 14/Jan/21 $$\mathrm{qn}\:\mathrm{128582}…
Question Number 63636 by naka3546 last updated on 06/Jul/19 $${Find}\:\:{all}\:\:{solutions}\:\:{of}\:\:\left({x},\:{y},\:{a},\:{b}\right)\:\:{for}\:\:{these}\:\:{equations}\:: \\ $$$$\:\:\:\:\:\:\:\:{x}\:+\:{y}^{\mathrm{2}} \:\:=\:\:{p}^{{a}} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:{y}\:\:=\:\:{p}^{{b}} \\ $$$${which}\:\:\:{x},\:{y},\:{a},\:{b}\:\:{are}\:\:{integers}\:\:{and}\:\:{p}\:\:{prime}\:\:{number}\:. \\ $$ Terms of Service Privacy Policy…
Question Number 129147 by Adel last updated on 13/Jan/21 $$\mathrm{cot}^{\mathrm{2}} \boldsymbol{\pi}/\mathrm{7}+\mathrm{cot}^{\mathrm{2}} \mathrm{2}\boldsymbol{\pi}/\mathrm{7}+\mathrm{cot}^{\mathrm{2}} \mathrm{3}\boldsymbol{\pi}/\mathrm{7}=? \\ $$ Commented by benjo_mathlover last updated on 13/Jan/21 $$\:\mathrm{ans}\::\:\mathrm{5} \\ $$…
Question Number 129130 by bounhome last updated on 13/Jan/21 $${solve}\:: \\ $$$$\:\:{y}''+\mathrm{4}{y}={cos}\mathrm{2}{x} \\ $$$$\: \\ $$ Answered by liberty last updated on 13/Jan/21 $$\:\mathrm{Homogenous}\:\mathrm{solution}\:\mathrm{y}_{\mathrm{h}} =\mathrm{C}_{\mathrm{1}}…
Question Number 129129 by Adel last updated on 13/Jan/21 Commented by MJS_new last updated on 13/Jan/21 $$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{e}}\:\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{yet}\:\mathrm{show}\:\mathrm{it} \\ $$ Commented by Adel last updated on…
Question Number 129098 by Adel last updated on 12/Jan/21 Answered by liberty last updated on 12/Jan/21 $$\mathrm{let}\:\mathrm{x}−\mathrm{1}=\mathrm{t} \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{t}}\:−\mathrm{sin}\:\mathrm{t}−\mathrm{2cos}\:\mathrm{t}}{\mathrm{arctan}\:\mathrm{t}−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{t}\right)}\:= \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\right)−\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}}…