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1-pi-1-22-7-1-3-1-7-1-3-1-1-7-1-1-7-let-1-pi-t-1-1-7-t-1-1-7-t-1-t-7-1-t-7-1-t-7-hence-t-1-1-pi-1-i-request-all-math-profe

Question Number 110197 by redmiiuser last updated on 27/Aug/20 $$\left(−\mathrm{1}\right)^{\pi} =? \\ $$$$\left(−\mathrm{1}\right)^{\frac{\mathrm{22}}{\mathrm{7}}} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{3}} .\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$=−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$${let}\:\left(−\mathrm{1}\right)^{\pi} ={t} \\ $$$$\Rightarrow−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}}…

if-y-ln-tan-4-x-2-show-that-dy-dx-secx-

Question Number 44622 by mondodotto@gmail.com last updated on 02/Oct/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{{y}}=\boldsymbol{\mathrm{ln}}\left[\boldsymbol{\mathrm{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\right]\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\boldsymbol{\mathrm{sec}{x}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18 $$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}×{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{cos}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}}…

given-that-sin-1-x-sin-1-y-c-show-that-dy-dx-1-y-2-1-x-2-0-

Question Number 44623 by mondodotto@gmail.com last updated on 02/Oct/18 $$\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{{x}}+\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{{y}}=\boldsymbol{\mathrm{c}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\sqrt{\frac{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}=\mathrm{0} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18…