Power series for sin(x)=x-x^3/6+x^5/120-... and the power series for tan(x)=x+x^3/3+2x^5/15+...
So, x/sin(x) is approximately 1/(1-x^2/6) and for tan(x)/x is approximately 1+x^2/3 when x is small.
1/(1-x^2/6)=1+x^2/6 approximately. Therefore, the difference is x^2/3-x^2/6=x^2/6>0.
This implies that for small x, tan(x)/x>x/sin(x).
[The expansions of sin, cos and tan as power series can be obtained through calculus and repeated differentiation, with sin and cos easily calculable and tan following from deriving sin and cos (or other method).
If necessary, I can include the derivation here. Please let me know if you wish it to be added to this answer.]
So, when x is near zero, tan(x)/x>x/sin(x). When x is close to (pi)/2, x/sin(x) approaches (pi)/2 while tan(x)/x approaches infinity. At x=(pi)/4, tan(x)/x=4/(pi)=1.27 approx, and x/sin(x)=(pi)sqrt(2)/4=1.11<1.27. With increasing values of x the difference between the two terms continues to increase, demonstrating that tan(x)/x>x/sin(x) over the specified range. The gradient of tan(x)/x-x/sin(x) also increases from near zero to approaching infinity as x moves between zero and (pi)/2.