How to simplify elegantly $\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right)$? I currently try to simplify the following trigonometric expression:
$$
\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right)
$$
where $t\in(0;1]$. 
I know that $\arctan(x)=\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)$ and I am also aware that there are formulas to simplify $\arcsin(x)+\arcsin(y)$, but they depend on different cases, as it can be seen here.
Is there a rather elegant way to solve the problem?
Any help is highly appreciated.
 A: Set
$$
f(t):=\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right), \qquad t \in (0,1]
$$ by differentiating
$$
f'(t)=0,\qquad t \in (0,1]
$$ thus $f$ is constant on $(0,1]$, putting  $t=1$ you get $f(1)=\dfrac\pi2$ giving

$$
\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right)=\frac\pi2 \qquad t \in (0,1]
$$

A: Let $\arctan\sqrt{\dfrac{1-t}t}=u$
Though the principal value of $\arctan$ lies in $\in\left[-\dfrac\pi2,\dfrac\pi2\right]$
As $\sqrt{\dfrac{1-t}t}\ge0\implies0\le u\le\dfrac\pi2\iff0\le2u\le\pi$
and $\dfrac{1-t}t=\tan^2u\implies t=\cos^2u$
$\implies\arcsin(2t-1)=\arcsin(2\cos^2u-1)=\arcsin\cos2u=\dfrac\pi2-\arccos(\cos2u)$
Now $\arccos(\cos2u)=\begin{cases} 2u &\mbox{if } 0\le2u\le\pi \\ 
-2u & \mbox{if } -\pi\le2u<0\end{cases}$
Hope you can take it from here.
A: $\displaystyle \sin^{-1}(2t-1)+2\tan^{-1}\left(\sqrt{\frac{1-t}{t}}\right)\;,$ Where $0<t\leq 1$
Now Using $\displaystyle \sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}\Rightarrow \sin^{-1}(x) = \frac{\pi}{2}-\cos^{-1}(x)$
Now expression is $\displaystyle \frac{\pi}{2}-\cos^{-1}(2t-1)+2\tan^{-1}\left(\sqrt{\frac{1-t}{t}}\right)$
Now Put $t=\cos^2 \phi\;,$ Then $0<\cos^2 \phi \leq 1\Rightarrow -1 \leq \cos \phi \leq 1-\left\{0\right\}$
So we get $\displaystyle  = \frac{\pi}{2}-\cos^{-1}\left(2\cos^2 \phi -1 \right)+2\tan^{-1}\left(\sqrt{\frac{1-\cos^2 \phi}{\sin^2 \phi}}\right)$
So we get $ \displaystyle = \frac{\pi}{2}-\cos^{-1}(\cos 2\phi)+2\tan^{-1}\left(|\tan \phi|\right)\;,$ Where $\displaystyle 0 \leq \phi \leq \pi-\left\{\frac{\pi}{2}\right\}$
A: Here is a nice and fun (maybe elegant?) approach.  For  $\frac{1}{2}<t\leq 1$ consider the Figure below. 
$\hskip1.5in$ 
$\triangle ABC$ is a right-angled triangle with side $\overline{BC} = 2t-1$ and hypotenuse $\overline{AC} = 1$, so that
$$\alpha = \arcsin(2t-1).$$
By Pythagorean Theorem, $\overline{AB} = 2\sqrt{t-t^2}.$ Extend $BC$ to a segment $\overline{BD} = 2t$, so that $\overline{CD} = 1$ and
$$\beta = \arctan\sqrt{\frac{1-t}{t}}.$$
Using the fact that $\triangle ACD$ is isosceles and $\triangle ABD$ is right-angled we can write
$$(\beta + \alpha) + \beta = \frac{\pi}{2},$$
i.e.
\begin{equation}\arcsin(2t-1) +  2\arctan\sqrt{\frac{1-t}{t}} = \frac{\pi}{2}.\tag{1}\label{1}\end{equation}
For $0<t<\frac{1}{2}$ consider the following Figure.
$\hskip1.5in$ 
Now $\overline{BC} = -2t+1$, and $\overline{AC} = 1$,
so that
$$\alpha = -\arcsin(2t-1).$$
Again we have $\overline{AB} = 2\sqrt{t-t^2}$. Extend $BC$ to a segment $\overline{CD} = 1$, so that $\overline{BD} = 2$ and
$$\beta = \arctan\sqrt{\frac{1-t}{t}}.$$
We obtain again \eqref{1} by considering realtionships between $\alpha$ and $\beta$ due to the fact that $\triangle ACD$ is isosceles, i.e.
$$\beta +(\beta - \alpha) = \frac{\pi}{2}.$$
$\blacksquare$
