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`        `
7 years ago

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

```
7 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

```
7 years ago
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