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Mohit Mittal Grade: 12
        

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6 years ago

Answers : (2)

Askiitians_Expert Yagyadutt
askIITians Faculty
74 Points
										

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


--------------------------------------------------------------------------------------------------------------------------------------


 


 


 

6 years ago
rajan jha
49 Points
										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?

6 years ago
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