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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))
ify=1+sinx1sinxshowthatdydx=11sinx
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=2tanx21+tan2x2y=1+2tanx21+tan2x212tanx21+tan2x2y=(1+tanx2)2(1tanx2)2y=1+tanx21tanx2=tan(Π4+x2)dydx=sec2(Π4+x2).12dydx=12cos2(Π4+x2)=11+cos{2.(Π4+x2)}=11+cos(Π2+x)dydx=11sinxproved
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=(1+sinx)(1sinx)(1sinx)2y=cosx1sinxdydx=sinx(1sinx)cosx.(cosx)(1sinx)2dydx=sinx+sin2x+cos2x(1sinx)2=1sinx(1sinx)2dydx=11sinx
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))
Itcanbeshownthaty=1+sinxcosxy=secx+tanxdydx=secxtanx+sec2x=secx(secx+tanx)=1+sinxcos2x=11sinx

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