Given curve is y = (1+x)y + sin−1(sin2x) On differentiating w.r.t x, we get
dy
dx
= (1+x)y[
y
1+x
+log(1+x)
dy
dx
] +
2sinxcosx
√1−sin4x
⇒ (
dy
dx
)at(0,1) = 1 [Since at x = 0 , y = 1] Slope of normal at (x = 0) = -1 ∴ Equation of normal at x = 0 and y = 1 is y - 1 = - 1 (x - 0) ⇒ y - 1 = - x ⇒ x + y = 1