I could show in calculus where $0<θ<\frac{\pi}{2}$ like this
$$\frac{d(θ)}{dθ}=1<\frac{1}{2}(\cosθ+\sec^2θ)$$
But I curioused about would it could prove by geometric way likewise trigonometric identities. I try to show with the area of circular sector but I failed...
So this is question:
How to show $$\frac{\sinθ +\tanθ}{2}>\theta$$ where $0<\theta<\frac{\pi}{2}$ in geometric way?
Here are some thoughts:
Approach $1a$
The blue region is a unit arc with an angle $\theta$. Let the length of $\overline {IH}$ be $\displaystyle L=\frac{\sin\theta+\tan\theta}2$.
The area of $\triangle AIH$ is given by $$S=\frac12 L\left(\frac L{\tan\theta}\right)=\frac{L^2}{2\tan\theta}\tag{1}$$
Notice the area of $\triangle ADG> \text{arc } ACG\implies\tan\theta>\theta\tag 2$
By $(1), (2)$
$$S<\frac{L^2}{2\theta}$$
Approach 1b
Notice arc $CJG$ has the length $\theta$. Then to prove $\theta<L$ is equivalent to prove the length of arc $CJG$ is shorter than the length of line $\overline {IH}$.