Answer
The equation of the tangent line $(l)$ to $y$ at point $(0,1)$ would be $$(l): y=x+1$$
Work Step by Step
$$y=\sin x+\cos x$$
1) Find the derivative of $y$ $$y'=(\sin x+\cos x)'$$ $$y'=\cos x-\sin x$$
2) Find $y'(0)$ $$y'(0)=\cos 0-\sin0$$ $$y'(0)=1-0=1$$
3) We know $y'(0)$, therefore the slope of the tangent line $(l)$ to $y$ at point $A(0,1)$ is also known, since it equals $y'(0)$.
So the equation of the tangent line $(l)$ to $y$ at point $A(0,1)$ would be $$(l): y-1=y'(0)(x-0)$$ $$(l): y-1=1\times x$$ $$(l): y=x+1$$
