Question
If $$\theta $$ is a positive acute angle then :
A.
$$\tan \,\theta < \theta < \sin \,\theta $$
B.
$$\theta < \sin \,\theta < \tan \,\theta $$
C.
$$\sin \,\theta < \tan \,\theta < \theta $$
D.
none of these
Answer :
none of these
Solution :
$$\eqalign{
& {\text{Let }}f\left( \theta \right) = \tan \,\theta - \theta \cr
& {\text{Then }}f'\left( \theta \right) = {\sec ^2}\theta - 1 = {\tan ^2}\theta > 0 \cr} $$
$$\therefore \,f\left( \theta \right)$$ is increasing. Therefore, $$f\left( \theta \right) > f\left( 0 \right)\,\,\,\,\, \Rightarrow \tan \,\theta - \theta > 0$$
Let $$\phi \left( \theta \right) = \theta - \sin \,\theta $$
Then $$\phi '\left( \theta \right) = 1 - \cos \,\theta = 2{\sin ^2}\frac{\theta }{2} > 0$$
$$\therefore \,\phi \left( \theta \right)$$ is increasing. Therefore, $$\phi \left( \theta \right) > \phi \left( 0 \right)\,\, \Rightarrow \sin \,\theta - \,\theta > 0$$