Author: Tinku Tara

Question Number 224739 by thetpainghtun_111 last updated on 30/Sep/25 $$\mathrm{z}\:=\:\mathrm{r}\:\left(\mathrm{cos}\:\theta\:+\:\mathrm{i}\:\mathrm{sin}\:\theta\right),\:\mathrm{find}\:\frac{\mathrm{z}}{\overset{−} {\mathrm{z}}}\:+\frac{\overset{−} {\mathrm{z}}}{\mathrm{z}}. \\ $$ Answered by Frix last updated on 30/Sep/25 $${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Leftrightarrow\:\bar {{z}}={r}\mathrm{e}^{−\mathrm{i}\theta} \\…

Question Number 224714 by fantastic last updated on 28/Sep/25 $$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\theta}}\:{d}\theta \\ $$ Commented by Ghisom_ last updated on 29/Sep/25 $$\int\frac{{d}\theta}{\:\sqrt{\mathrm{tan}\:\theta}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:\theta}\:\rightarrow\:{d}\theta=\mathrm{2cos}^{\mathrm{2}} \:\theta\:\sqrt{\mathrm{tan}\:\theta}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}}…

Question Number 224692 by TonyCWX last updated on 27/Sep/25 $$ \\ $$ Commented by TonyCWX last updated on 27/Sep/25 Commented by TonyCWX last updated on…

Question Number 224678 by fantastic last updated on 26/Sep/25 $$\left.\mathrm{Guys}\:\mathrm{I}\:\mathrm{just}\:\mathrm{turned}\:\mathrm{15}\:\mathrm{today}\::\right)\: \\ $$ Commented by mingski last updated on 26/Sep/25 $$\mathrm{happy}\:\mathrm{birthday}! \\ $$ Commented by Rasheed.Sindhi…