Answer:
Option A
Explanation:
Given curve is
$y-(1+x)^{y}+ \sin^{-1} (\sin^{2}x)$
On differentiating w.r.t x , we get
$\frac{dy}{dx}=(1+x)^{y}\left[\frac{y}{1+x}+\log (1+x)\frac{dy}{dx}\right]+\frac{2 \sin x \cos x}{\sqrt{1-\sin^{4}x }}$
$\Rightarrow$ $\left(\frac{dy}{dx}\right)_{at(0,1)}=1 [ \because at x=0,y=1]$
Slope of normal at (x=0)=-1
$\therefore$ Equation of normal at x=0 and y=1 is y-1=-1(x-0)
$\Rightarrow$ $y-1=-x \Rightarrow x+y=1$
