One of my friends gave me this problem. Initially I thought that this would be easy but i tried this for two days unable to do this in a proper manner.
There is a circle with an angle A located at the center of the circle
Prove that: $\sin A < A < \tan A$
The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$.
By inclusion, we get
$$
\frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1}
$$
On $(0,\pi/2)$ we have $\cos t < 1 < \sec^2 t.$ Integrating each term from $t=0$ to $t=A$ gives your desired inequality as long as $0<A<\pi/2.$ [inequality false for at least some $A$ not in this range.]
Of course this proof is usually done via a diagram; see a calc book.
The shortest way between two points is the straight line. This establishes $$\sin \dfrac A2<\dfrac A2.$$
Then the area of the triangle formed with the tangent to the circle is larger than the area of the sector, hence
$$\frac A2<\tan\frac A2.$$
Replace $\dfrac A2$ by $A$.
