Let f (x)=sin x –tan x, x∈(0,π2), then tangent drawn to the curve in the given interval will lie;

f (x) = sin x –tan x ⇒f '(x) = cos x –sec2x ⇒f ''(x) = −sin x –2 sec2x. tan x =−(sinx+2sinxcos3x) =[−] ve for x∈R(0,   π2)  ⇒  Tangent drawn to the curve will lie above the curve.