The range of real constant t such that $(1-t)\sin x + t\tan x >x$ always holds for $x$ between $0$ and $\pi/2$ I converted the expression to 
$t>\frac{x-\sin x}{\tan x-\sin x}=f(x)$ so basically now all I require is the maximum value of $f(x)$ in the given interval.
I looked up the graph of $f(x)$ (on Desmos) and it showed $f(x)$ is decreasing between $0$ and $\pi/2$. 
The maximum value of $f(x)$ comes out to be $f(x)$ when $x$ tends to zero, and it matches the answer perfectly.
So basically the question reduces to proving that $f(x)$ is decreasing in the interval $\left(0,\frac{\pi}{2}\right).$
It would be great if somebody could help out.
 A: Let us set
$$ f_t(x) = (1-t)\sin x+t\tan x-x \tag{1}$$
and consider the values of $t$ in the range $[0,1]$ first. We have $\lim_{x\to{\frac{\pi}{2}}^-}f_t(x)=+\infty$ due to the singularity of the tangent function and 
$$ f_t(x) = \left(\frac{t}{2}-\frac{1}{6}\right)x^3+\left(\frac{1}{120}+\frac{t}{8}\right)x^5+\ldots \tag{2}$$
in a right neighbourhood of the origin, by well-known Taylor series. We may notice at once that if $t<\frac{1}{3}$ then $f_t(x)$ is negative in a right neighbourhood of the origin. Then we may move to prove the following

Claim. For any $t\in\left[\frac{1}{3},1\right]$ we have $f_t(x)>0$ for any $x\in\left(0,\frac{\pi}{2}\right)$.

Proof. It is well known that all the odd derivatives of the tangent function at the origin are positive numbers. In particular the Taylor coefficients appearing in $(2)$  are increasing functions with respect to $t$. The claim with $t=\frac{1}{3}$ is known as Huygens' inequality, and since the claim holds for $t=\frac{1}{3}$ it holds for any $t\in\left[\frac{1}{3},1\right]$
It remains to study the cases $t<0$ and $t>1$. They are pretty simple, hence I leave this task to you.
