Function f (x) = sin x + tan x + sgn (x^2-6x+10) is

$sgn (x^2-6x+10)=1=sgn(x^2-2×3x+9+1)$

$⇒sgn((x-3)^2+1)=1$

as $(x-3)^2≥0$

$(x-3)^2+1>0$

so $f(x)=\sin x + \tan x + 1$

as period of $\sin x$ is 2π

period of $\tan x$ is 2π

Period is 2π