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Question Number 41536 by mondodotto@gmail.com last updated on 09/Aug/18
if y=(√(((1+sinx)/(1−sinx)) )) show that (dy/dx)=(1/(1−sinx))
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{{y}}=\sqrt{\frac{\mathrm{1}+\boldsymbol{\mathrm{sin}{x}}}{\mathrm{1}−\boldsymbol{\mathrm{sin}{x}}}\:}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{\mathrm{sin}{x}}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
sinx=((2tan(x/2))/(1+tan^2 (x/2))) y=(√((1+((2tan(x/2))/(1+tan^2 (x/2))))/(1−((2tan(x/2))/(1+tan^2 (x/2)))))) y=(√(((1+tan(x/2))^2 )/((1−tan(x/2))^2 ))) y=((1+tan(x/2))/(1−tan(x/2)))=tan((Π/4)+(x/2)) (dy/dx)=sec^2 ((Π/4)+(x/2)).(1/2) (dy/dx)=(1/(2cos^2 ((Π/4)+(x/2))))=(1/(1+cos{2.((Π/4)+(x/2))}))=(1/(1+cos((Π/2)+x))) (dy/dx)=(1/(1−sinx)) proved
$${sinx}=\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$${y}=\sqrt{\frac{\mathrm{1}+\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}{\mathrm{1}−\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}}\: \\ $$$${y}=\sqrt{\frac{\left(\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }}\: \\ $$$${y}=\frac{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}}={tan}\left(\frac{\Pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right) \\ $$$$\frac{{dy}}{{dx}}={sec}^{\mathrm{2}} \left(\frac{\Pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right).\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\Pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{1}+{cos}\left\{\mathrm{2}.\left(\frac{\Pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)\right\}}=\frac{\mathrm{1}}{\mathrm{1}+{cos}\left(\frac{\Pi}{\mathrm{2}}+{x}\right)} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}−{sinx}}\:{proved} \\ $$
Answered by math1967 last updated on 09/Aug/18
y=(√(((1+sinx)(1−sinx))/((1−sinx)^2 ))) y=((cosx)/(1−sinx)) (dy/dx)=((−sinx(1−sinx)−cosx.(−cosx))/((1−sinx)^2 )) (dy/dx)=((−sinx+sin^2 x+cos^2 x)/((1−sinx)^2 ))=((1−sinx)/((1−sinx)^2 )) (dy/dx)=(1/(1−sinx))
$${y}=\sqrt{\frac{\left(\mathrm{1}+{sinx}\right)\left(\mathrm{1}−{sinx}\right)}{\left(\mathrm{1}−{sinx}\right)^{\mathrm{2}} }} \\ $$$${y}=\frac{{cosx}}{\mathrm{1}−{sinx}} \\ $$$$\frac{{dy}}{{dx}}=\frac{−{sinx}\left(\mathrm{1}−{sinx}\right)−{cosx}.\left(−{cosx}\right)}{\left(\mathrm{1}−{sinx}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{−{sinx}+{sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{sinx}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−{sinx}}{\left(\mathrm{1}−{sinx}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}−{sinx}} \\ $$$$ \\ $$
Answered by $@ty@m last updated on 09/Aug/18
It can be shown that y=((1+sin x)/(cos x)) y=sec x+tan x (dy/dx)=sec xtan x+sec^2 x =sec x(sec x+tan x) =((1+sin x)/(cos^2 x)) =(1/(1−sin x))
$${It}\:{can}\:{be}\:{shown}\:{that} \\ $$$${y}=\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$${y}=\mathrm{sec}\:{x}+\mathrm{tan}\:{x} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{sec}\:{x}\mathrm{tan}\:{x}+\mathrm{sec}\:^{\mathrm{2}} {x} \\ $$$$=\mathrm{sec}\:{x}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right) \\ $$$$=\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:{x}} \\ $$

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