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If-cosec-1-x-cot-1-y-show-that-d-2-y-dx-2-1-y-3-0-




Question Number 117482 by Don08q last updated on 12/Oct/20
 If  cosec^(−1) x  =  cot^(−1) y, show that                        (d^2 y/dx^2 )  +  (1/y^3 )  =  0
Ifcosec1x=cot1y,showthatd2ydx2+1y3=0
Answered by TANMAY PANACEA last updated on 12/Oct/20
y=cot(cosec^(−1) x)  θ=cosec^(−1) x→cosecθ=x  x^2 =1+cot^2 θ→cotθ=(√(x^2 −1))   y=(√(x^2 −1))   (dy/dx)=(1/(2(√(x^2 −1))))×2x=(x/( (√(x^2 −1))))  (d^2 y/dx^2 )=(((√(x^2 −1)) ×1−x×(1/(2(√(x^2 −1))))×2x)/(x^2 −1))  =((x^2 −1−x^2 )/((x^2 −1)^(3/2) ))=((−1)/y^3 ) proved
y=cot(cosec1x)θ=cosec1xcosecθ=xx2=1+cot2θcotθ=x21y=x21dydx=12x21×2x=xx21d2ydx2=x21×1x×12x21×2xx21=x21x2(x21)32=1y3proved
Commented by $@y@m last updated on 12/Oct/20
Nice. I just want to add that there is no use of writing first line.
Commented by TANMAY PANACEA last updated on 13/Oct/20
thank you sir
thankyousir
Commented by Don08q last updated on 13/Oct/20
Thank you Sir
ThankyouSir
Answered by $@y@m last updated on 13/Oct/20
 Let cosec^(−1) x  =  cot^(−1) y=θ  x=cosec θ                    ∣  y=cot θ  (dx/dθ)=−cosec θcot θ    ∣ (dy/dθ)=−cosec^2 θ  (dy/dx)=((cosec θ)/(cot θ))  (dy/dx)= sec θ .....(1)  (d^2 y/dx^2 ) =(d/dx)(sec θ)=(d/dθ)(sec θ)×(dθ/dx)  (d^2 y/dx^2 ) =sec θtan θ×(1/(−cosec θcot θ))  (d^2 y/dx^2 ) =(1/(−cot^3 θ))=−(1/y^3 )  (d^2 y/dx^2 )+(1/y^3 )=0
Letcosec1x=cot1y=θx=cosecθy=cotθdxdθ=cosecθcotθdydθ=cosec2θdydx=cosecθcotθdydx=secθ..(1)d2ydx2=ddx(secθ)=ddθ(secθ)×dθdxd2ydx2=secθtanθ×1cosecθcotθd2ydx2=1cot3θ=1y3d2ydx2+1y3=0
Commented by Don08q last updated on 13/Oct/20
Thank you Sir
ThankyouSir

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