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Category: Algebra

Prove-that-tan-sec-1-tan-tan-1-cot-

Question Number 10542 by FilupS last updated on 17/Feb/17 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{tan}\left(\mathrm{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{tan}\left(\theta\right)}\right)\right)=\sqrt{\mathrm{tan}\left(\theta\right)}\sqrt{\mathrm{1}−\mathrm{cot}\left(\theta\right)} \\ $$ Answered by mrW1 last updated on 17/Feb/17 $${let}\:\alpha=\mathrm{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{tan}\:\left(\theta\right)}\right) \\…

1-k-1-C-n-k-

Question Number 141614 by ArielVyny last updated on 21/May/21 $$\Sigma\frac{\mathrm{1}}{{k}+\mathrm{1}}{C}_{{n}} ^{{k}} \:.\: \\ $$ Commented by Dwaipayan Shikari last updated on 21/May/21 $$\frac{\mathrm{1}}{{k}+\mathrm{1}}\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}{C}_{{n}}…

What-s-the-minimum-value-of-13a-13b-2c-2a-2b-24a-b-13c-2b-2c-a-24b-13c-2a-2c-a-b-c-are-positive-numbers-I-think-nobody-can-solve-this-

Question Number 76061 by tw000001 last updated on 23/Dec/19 $$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{13}{a}+\mathrm{13}{b}+\mathrm{2}{c}}{\mathrm{2}{a}+\mathrm{2}{b}}+\frac{\mathrm{24}{a}−{b}+\mathrm{13}{c}}{\mathrm{2}{b}+\mathrm{2}{c}}+\frac{−{a}+\mathrm{24}{b}+\mathrm{13}{c}}{\mathrm{2}{a}+\mathrm{2}{c}}? \\ $$$$\left({a},{b},{c}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{numbers}.\right) \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{nobody}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}. \\ $$ Answered by MJS last updated on 23/Dec/19…

e-x-2-dx-

Question Number 76048 by peter frank last updated on 22/Dec/19 $$\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$ Answered by Crabby89p13 last updated on 23/Dec/19 $${Actually}\:{there}\:{is}\:{no}\:{elementary} \\ $$$${function}\:{that}\:{describes}\:{the}\:{integral}\:{of}…

1-1-x-dx-

Question Number 141578 by Khalmohmmad last updated on 20/May/21 $$\int\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}{dx}=? \\ $$ Answered by bramlexs22 last updated on 20/May/21 $$\:\int\:\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}}\:{dx}\: \\ $$$$\:{let}\:\sqrt{\mathrm{1}+{x}}\:=\:{u}\:\Rightarrow{x}={u}^{\mathrm{2}} −\mathrm{1}\: \\ $$$${dx}\:=\:\mathrm{2}{u}\:{du}\:…

1-2-2-3-3-4-4-5-17-18-

Question Number 10493 by ABD last updated on 14/Feb/17 $$\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+\frac{\mathrm{4}}{\mathrm{5}!}+…+\frac{\mathrm{17}}{\mathrm{18}!}=? \\ $$ Answered by mrW1 last updated on 14/Feb/17 $${since}\:\frac{{n}}{\left({n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{{n}!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+\frac{\mathrm{4}}{\mathrm{5}!}+…+\frac{\mathrm{17}}{\mathrm{18}!} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+\centerdot\centerdot\centerdot+\left(\frac{\mathrm{1}}{\mathrm{17}!}−\frac{\mathrm{1}}{\mathrm{18}!}\right) \\…