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A-B-pi-4-find-the-1-tanA-1-tanB-with-explanotory-solution-

Question Number 186590 by mustafazaheen last updated on 06/Feb/23 $$\mathrm{A}+\mathrm{B}=\frac{\pi}{\mathrm{4}}\:\:\:\:\:{find}\:{the}\:\left(\mathrm{1}+\mathrm{tan}{A}\right)\left(\mathrm{1}+\mathrm{tan}{B}\right)=?\: \\ $$$${with}\:{explanotory}\:{solution} \\ $$ Answered by mr W last updated on 06/Feb/23 $$\mathrm{tan}\:\left({A}+{B}\right)=\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\mathrm{1} \\ $$$$\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{B}}{\mathrm{1}−\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}}=\mathrm{1}…

Find-intgral-of-tan-x-by-parts-tan-x-dx-pls-help-me-i-am-a-newbie-in-calculus-

Question Number 55493 by Kunal12588 last updated on 25/Feb/19 $${Find}\:{intgral}\:{of}\:\mathrm{tan}\:{x}\:{by}\:{parts} \\ $$$$\int{tan}\:{x}\:{dx}\:\:\:{pls}\:{help}\:{me}\:\:\left({i}\:{am}\:{a}\:{newbie}\:{in}\:{calculus}\right) \\ $$ Commented by Kunal12588 last updated on 25/Feb/19 $${i}\:{am}\:{getting}\:{something}\:{like}\:{this} \\ $$$$\int\left(\frac{\mathrm{1}}{{cosx}}\right)×\left({sin}\:{x}\:{dx}\right) \\…

Question-186555

Question Number 186555 by yaslm last updated on 05/Feb/23 Commented by yaslm last updated on 06/Feb/23 I am sorry, I need domain and Draw D Answered by qaz last updated on 06/Feb/23 $$\left[{Df}\right]=\begin{bmatrix}{\frac{\partial{f}}{\partial{x}}}&{\frac{\partial{f}}{\partial{y}}}\end{bmatrix}=\begin{bmatrix}{\mathrm{tan}\:{y}}&{{x}\mathrm{sec}\:^{\mathrm{2}}…

1-lim-x-x-1-x-x-2-x-2-x-2-x-dx-2-2-xcos-2x-1-2x-cos-3x-dx-3-n-5-4n-3n-2n-

Question Number 186539 by test1234 last updated on 05/Feb/23 $$\left(\mathrm{1}\right)\int\left(\underset{{x}\rightarrow{x}+\mathrm{1}} {\mathrm{lim}}\frac{{x}^{{x}+\mathrm{2}} {x}^{\mathrm{2}+{x}^{\mathrm{2}} } }{\:\sqrt{{x}}}\right){dx}=? \\ $$$$\left(\mathrm{2}\right)\int\frac{\mathrm{2}+{x}\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{2}{x}+\mathrm{cos}\:\mathrm{3}{x}}{dx}=? \\ $$$$\left(\mathrm{3}\right)\:\underset{{n}=\mathrm{5}} {\overset{\mathrm{4}{n}} {\sum}}\left(\mathrm{3}{n}+\mathrm{2}{n}\right)=? \\ $$ Terms of Service…

Question-120982

Question Number 120982 by Khalmohmmad last updated on 04/Nov/20 Answered by Dwaipayan Shikari last updated on 04/Nov/20 $$\underset{{k}=\mathrm{2}} {\overset{\mathrm{625}} {\sum}}{log}_{\mathrm{5}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}}\right) \\ $$$$={log}_{\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}…..\frac{\mathrm{624}}{\mathrm{625}}\right)=−\mathrm{4} \\…