Question Number 20703 by tammi last updated on 01/Sep/17 $$\frac{\mathrm{sin}\alpha+\mathrm{sin}\:\mathrm{3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\:\alpha+\mathrm{cos}\:\mathrm{3}\alpha+\mathrm{cos}\:\mathrm{5}\alpha}=\mathrm{tan}\:\mathrm{3}\alpha \\ $$ Commented by myintkhaing last updated on 01/Sep/17 $$\mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{\mathrm{2sin3}\alpha\mathrm{cos2}\alpha+\mathrm{sin3}\alpha}{\mathrm{2cos3}\alpha\mathrm{cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{sin3}\alpha}{\mathrm{cos3}\alpha}=\mathrm{tan3}\alpha \\ $$$$\mathrm{proved} \\ $$ Answered…
Question Number 20702 by tammi last updated on 01/Sep/17 $$\mathrm{tan}\:\mathrm{20}°\mathrm{tan}\:\mathrm{40}°\mathrm{tan}\:\mathrm{80}°=\sqrt{\mathrm{3}} \\ $$ Answered by Tinkutara last updated on 01/Sep/17 $$\mathrm{Using}\:\mathrm{tan}\:\theta\mathrm{tan}\:\left(\mathrm{60}−\theta\right)\mathrm{tan}\:\left(\mathrm{60}+\theta\right)=\mathrm{tan}\:\mathrm{3}\theta \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{20tan}\:\mathrm{40tan}\:\mathrm{80}=\mathrm{tan}\:\mathrm{60}=\sqrt{\mathrm{3}} \\ $$ Terms…
Question Number 151768 by mathdanisur last updated on 22/Aug/21 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Aug/21…
Question Number 86230 by M±th+et£s last updated on 27/Mar/20 $${is}\:\:\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} \:=\left(\sqrt[{{n}}]{\left(−\mathrm{1}\:\right)^{{m}} }\right)\:{or}\:=\left(\sqrt[{{n}}]{−\mathrm{1}}\right)^{{m}} \\ $$$${or}\:{both}\:{of}\:{them}\:{are}\:{fault}\:{and}\:{why}\:? \\ $$ Answered by MJS last updated on 27/Mar/20 $$\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} =\left(\mathrm{e}^{\mathrm{i}\pi}…
Question Number 86231 by liki last updated on 27/Mar/20 Commented by liki last updated on 27/Mar/20 $$..{plz}\:{i}\:{need}\:{help}\:{qn}\:{no}.\:\mathrm{5}\left({a}\right)\:\&\:{c} \\ $$ Answered by TANMAY PANACEA. last updated…
Question Number 151767 by mathdanisur last updated on 22/Aug/21 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{k}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Aug/21 $$\mathrm{my}\:\mathrm{calculous}\:\mathrm{was}\:\mathrm{false}. \\…
Question Number 20693 by Tinkutara last updated on 31/Aug/17 $$\mathrm{Integers}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:….,\:{n},\:\mathrm{where}\:{n}\:>\:\mathrm{2},\:\mathrm{are} \\ $$$$\mathrm{written}\:\mathrm{on}\:\mathrm{a}\:\mathrm{board}.\:\mathrm{Two}\:\mathrm{numbers}\:{m},\:{k} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{1}\:<\:{m}\:<\:{n},\:\mathrm{1}\:<\:{k}\:<\:{n}\:\mathrm{are} \\ $$$$\mathrm{removed}\:\mathrm{and}\:\mathrm{the}\:\mathrm{average}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{remaining}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{found}\:\mathrm{to}\:\mathrm{be}\:\mathrm{17}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{removed}\:\mathrm{numbers}? \\ $$ Answered…
Question Number 151766 by mathdanisur last updated on 22/Aug/21 Commented by tabata last updated on 23/Aug/21 $${ln}\left({y}\right)\:{ln}\left({x}\right)={ln}\left(\mathrm{3}\right)\Rightarrow{ln}\left({y}\right)=\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left({x}\right)} \\ $$$$ \\ $$$$\Rightarrow\frac{{y}^{'} }{{y}}=−\frac{{ln}\left(\mathrm{3}\right)}{{x}\:\left({ln}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$…
Question Number 151761 by talminator2856791 last updated on 22/Aug/21 $$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{1}}{\:\sqrt{{e}^{{x}} +\mathrm{1}}}\:{dx}\: \\ $$$$\: \\ $$ Answered by mindispower last updated on…
Question Number 151756 by Integrals last updated on 22/Aug/21 Commented by MJS_new last updated on 22/Aug/21 $$\int\left(\mathrm{cot}^{\mathrm{1}/\mathrm{4}} \:\beta\:+\mathrm{tan}^{\mathrm{1}/\mathrm{4}} \:\beta\right){d}\beta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}^{\mathrm{1}/\mathrm{4}} \:\beta\:\rightarrow\:{d}\beta=\frac{\mathrm{4}{t}^{\mathrm{3}} }{{t}^{\mathrm{8}} +\mathrm{1}}{dt}\right] \\…