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Question Number 148825 by DELETED last updated on 31/Jul/21 Answered by DELETED last updated on 31/Jul/21 $$\Sigma\mathrm{IR}+\Sigma\mathrm{E}=\mathrm{0} \\ $$$$\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{i}_{\mathrm{3}} \:\rightarrow\mathrm{i}_{\mathrm{2}} \:=\mathrm{i}_{\mathrm{3}} −\mathrm{i}_{\mathrm{2}} \:……\left(\mathrm{1}\right)…

Question Number 148798 by abdurehime last updated on 31/Jul/21 Answered by puissant last updated on 31/Jul/21 $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\int\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}+\int\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)}}\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}+\sqrt{\mathrm{cotan}\left(\mathrm{x}\right)}\right)\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}}\right)\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{tan}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}}\mathrm{dx}…

Question Number 148783 by 0731619 last updated on 31/Jul/21 Commented by hknkrc46 last updated on 31/Jul/21 $$\left(\mathrm{1}\right)\:\underset{\boldsymbol{{h}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{f}}\left(\boldsymbol{{x}}\:+\:\boldsymbol{{h}}\right)\:−\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{h}}}\: \\ $$$$=\:\underset{\boldsymbol{{h}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{cos}\:\left(\boldsymbol{{x}}\:+\:\boldsymbol{{h}}\right)\:+\:\mathrm{cos}\:\boldsymbol{{x}}}{\boldsymbol{{h}}} \\ $$$$=\:\underset{\boldsymbol{{h}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\boldsymbol{{x}}\:−\:\mathrm{cos}\:\boldsymbol{{x}}\:\centerdot\:\mathrm{cos}\:\boldsymbol{{h}}\:+\:\mathrm{sin}\:\boldsymbol{{x}}\:\centerdot\:\mathrm{sin}\:\boldsymbol{{h}}}{\boldsymbol{{h}}} \\…

Question Number 148724 by Sozan last updated on 30/Jul/21 $${find}\:{laurent}\:{series}\:{f}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −{z}+\mathrm{1}}\:\:,\mathrm{0}<\mid{z}−\mathrm{1}\mid<\mathrm{1} \\ $$ Answered by mathmax by abdo last updated on 30/Jul/21 $$\mathrm{the}\:\mathrm{way}\:\mathrm{of}\:\mathrm{this}\:\mathrm{kind}\:\mathrm{is}\:\mathrm{to}\:\mathrm{use}\:\mathrm{changement}\:\mathrm{z}−\mathrm{1}=\mathrm{y}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\varphi\left(\mathrm{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}}…

Question Number 148723 by Rustambek last updated on 30/Jul/21 $${tgx}+{tgy}=\mathrm{4}\:\:\:{cosx}+{cosy}=\frac{\mathrm{1}}{\mathrm{5}}\:\:\: \\ $$$${tg}\left({x}+{y}\right)=? \\ $$ Commented by Rustambek last updated on 30/Jul/21 $${please} \\ $$ Terms…