Hello Mohit

 

Question : 2

 

x.root(1+y) + y.root(1+x) = 0

x / root(1+x) = -y/root(1+y)

square both side

 

x^2/(1+x) = y^2/(1+y)

 

x^2 + x^2.y = y^2 + x.y^2

 

x^2 - y^2 = xy^2 - x^2y

(x-y)(x+y) = xy(y-x)

so  x + y = -xy

or  x + y + xy = 0 

 

y(1+x) =- x

y = -x/(1+x)

 

dy/dx = -1/(1+x)^2  ans

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let x=sinb & y=sint then put it in the question. u got cosb + cost = a(sinb-sint) =)2cos(b+t/2).cos(b-t/2)=a(2cos(b+t/2).sin(b-t/2)) =)cos(b-t/2)=asin(b-t/2) (after cancelling 2cos(b+t/2) both sides) =)cot(b-t/2)=a =)b-t=2cot-1(a) (here -1 denotes inverse) =)sin-1(x)-sin-1(y)=2cot-1(a) differentiating both sides, and get the r.h.s as zero as it is constant next step the answer comes like my answer?