Vtyjkyuk

let $$f(x)=sin(x)+tan(x)-2x$$ then $$f(0)=0$$ and $$f'(x)=cos(x)+frac1cos(x)^2-2$$ and this can be written as $$f'(x)=frac(1-cos(x))(1-cos(x)^2)+cos(x)(1-cos(x))cos(x)^2>0$$

Here's a geometric argument, but it isn't as slick as some of the Calculus-based ones.

Consider the unit circle about $O$, through $R$ and $S$, with $theta = angle ROS$. The perpendicular from $S$ to $overlineOR$ has length $sintheta$, while the perpendicular from $R$ up to $T$ on the extension of $overlineOS$ has length $tantheta$. Let $M$ be the midpoint of $overlineST$.

Then
$$2;|textarea of sector;ROS| = theta qquadtextandqquad 2;|triangle ORM| = frac12left(sintheta + tanthetaright)$$

"All we need to do" is show that the triangle has more area than the sector. This seems pretty clear; after all, the triangle contains almost-all of the sector, except for the circular segment defined by $overlineKR$, where $K$ is the intersection of $overlineRM$ and the circle. There is a concern, though, that the excess area in the triangular region $KSM$ could be less than that of the tiny sliver of a circular segment for small $theta$; we need to dispel that concern.

There's probably a simpler route to this, but I coordinatized and, with the help of Mathematica, found
$$M = left(frac1 + costheta2, fracsintheta (1 + costheta)2 costhetaright)$$
$$K = left(frac1 + 3 costheta + 2 cos^2theta + 2 cos^3theta1 + 3 costheta + 4 cos^2theta, frac2 sintheta costheta ( 1 + costheta)1 + 3 costheta + 4 cos^2thetaright)$$
so that (after a bit more symbol-crunching)
$$frac = frac1 + 3 costheta4 cos^2theta = 1 + frac1 + 3 costheta - 4 cos^2theta4 cos^2theta = 1 + frac(1-costheta)(1 + 4 costheta)4 cos^2theta > 1$$

for $0 < theta < pi/2$.

This says that $overlineMK$ is longer than $overlineKR$, so that we could reflect $R$ in $K$ to get $R^prime$, and copy circular segment $KR$ as circular segment $KR^prime$ inside $triangle ORM$ yet tangent to the unit circle (and therefore outside of sector $ORS$).

Consequently, the triangle definitely has more area than the sector, so we're done. $square$

For any $x$ in the given interval

$$sin(x)+tan(x)=int_0^xleft(cos(u)+frac1cos^2(u)right),dustackrelAM-GMgeq2int_0^xfracdusqrtcos(u) $$
and the last integral is clearly $>2int_0^x1,dx = 2x$.

The same approach proves the stronger, non-trivial inequality

$$ forall xinleft(0,tfracpi2right),qquad colorred2sin(x)+tan(x)>colorred3x.$$

Here it is a properly geometric proof.

Given a circle sector with amplitude $2theta$, we may consider the associated arc and the parabola through the midpoint and the endpoints of such arc. The union of the parabolic segment and the triangle with side lengths $1,1,2sintheta$ is a region strictly contained in the circle sector. The area of the parabolic segment is $frac43$ of the area of the shaded triangle, hence we have:

$$costhetasintheta+frac43sintheta(1-costheta)< theta.$$

We have $tan xgt x$ for $xin(0,fracpi2)$ the result follows, Using AM-GM-HM inequalites we have ,
$$colorbluefracsin x+ tan x2 ge sqrtsin xtan x ge frac2frac1sin x +frac1tan x = 2tan frac x2 gt x$$

Indeed, $$ frac1sin 2u +frac1tan 2u= frac1sin 2u + fraccos 2u sin 2u =frac2cos^2 u2cossin u = frac1tan u$$

$$f=sin (x)+tan (x)-2 ximplies f'=cos (x)+sec ^2(x)-2$$ Now, using the tangent half-angle substitution, we have $$f'=fract^24+frac2sqrtt^2+4-1$$ Squaring leads to $$f'=0 qquad textif qquad -fract^616+fract^44+t^2=0$$ the real roots are $$t=pmsqrt2 left(1+sqrt5right)$$ Considering the positive root then $$f'=0 qquad textif qquad x=2tan ^-1left(sqrt2 left(1+sqrt5right)right)approx 2.39255$$ So, the derivative does not cancel in the interval and it is always non negative.

Funny would also be a Taylor expansion; up to any order, the coefficients are positive.

From $f' = tan^2 x-(1-cos x)$, we get $f'' = 2tan x sec^2 x - sin x = sin x(2sec^3 x -1) > 0$ in $left(0, fracpi2right)$. Thus $f'$ is increasing in $left(0, fracpi2right)$ and thus $f'(x) > f'(0) = 0$. Thus $f$ is increasing.

We've $dfrac(sin x + tan x)2ge sqrtsin xtan x$
(AM GM Inequality)

Equality holds when $x=0$

Using expansions of $sec x$ and $cos x$ one can easily show that $sin x tan x>x^2$

Hence the result

Here's a far cleaner argument than in my previous answer.

Considering acute (and non-zero) $theta$ ...

In the figure,
$overlinePS$ is a leg of a right triangle with hypotenuse $overlinePT$. Thus,
$$|PS| < |PT| quadimpliesquad |PR| < |PT| quadimpliesquad |PR| < frac12|TR| tag$star$$$
(FYI, this proves the convexity of the tangent function: $tanfrac12theta < frac12tantheta$.) Continuing, with "cseg" indicating "circular segment", ...

$$beginalign
|textcsegRS| ;<; |triangle PRS| &;stackrelstar<; frac12|triangle TRS| = frac12left(;|triangle TRO|-|triangle SRO|;right) \[6pt]
impliesquad|textsectRS| ;=; |textcsegRS| + |triangle SRO| &;<; frac12left(;|triangle TRO|+|triangle SRO|;right) \[6pt]
impliesquad frac12cdot 1^2cdottheta &;<; frac12left(;frac12cdot 1cdot tantheta ;+; frac12cdot 1cdotsintheta;right)
endalign$$

and the result follows. $square$

In fact one can prove a better inequality

$$ frac 2 sin x + tan x 3 > x$$ for $x in (0, fracpi2)$.

Indeed, the difference $frac 2 sin x + tan x 3 - x$ has derivative
$$frac13 cos^2 x ( 2 cos^3 x - 3 cos ^2 x + 1 )$$ and since the expression $2 t^3 - 3 t^2 + 1 = (t-1)^2 ( 2 t+ 1)$ is $>0$ on $[0,1)$, the function in $x$ is strictly increasing on $[0, pi/2]$.

Obs: One can check this is the "best" inequality of its kind.

$bfAdded:$ Like @Claude Leibovici: noticed for the original question, in this case also the Maclaurin expansion of the difference $2 sin x + tan x - 3 x$ has all the coefficients positive. This can be checked by using the explicit Taylor- Maclaurin expansions of $sin$ and $tan$.

Let $AB=AC=1$ and $angle BAC = alpha$. Let $D in AC$ with $DB perp AB$. Let $E in AB$ with $CE perp AB$. Let $F$ and $G$ be the midpoints of $EB$ and $CD$, respectively. Let $H in BD$ with $CH perp AD$. Let $I in FG$ with $IH perp GF$.

Clearly $IH < GD = CG$, therefore
$$IG^2 = GH^2 - IH^2 > GH^2 - CG^2 = CH^2.$$
Therefore $GI>CH$. It follows that
beginalign*
fracsin alpha + tan alpha2 & = fracCE+BD2 = GF = GI+IF
> CH+IF = CH+BH = 2BH \
&= 2tan frac alpha2 > 2 cdot frac alpha2 = alpha.endalign*