The correct option is C III quadrant
Let f(x) = tan x -x
When 0 < x < π2, tan x > x ⇒ f(x) > 0
∴f(x)=0 has no root in I quadrant
In the II and IV quadrant tan x is negative
∴ f(x) < 0 and f(x) = 0 has no solution in these quadrants
Next f(π)=tanπ−π <0
f(x)→∞asx→3π2
∴f(x)=0 has at least one root in the III quadrant.