How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
 A: Hint:
set $b=\cos t$:
\begin{align}
\sqrt{2+\cos t+\sqrt3\sqrt{1-\cos^2t}} 
&\overset?= 
\sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sqrt{1-\cos^2t}} 
,\\
\sqrt{2+\cos t+\sqrt3\sin t} 
&\overset?= 
\sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sin t} 
,\\
\sqrt{1+\tfrac12\cos t+\tfrac{\sqrt3}2\sin t} 
&\overset?=
\sqrt{\vphantom{\tfrac{\sqrt3}2}1-\cos t}+ \sqrt{1+\tfrac12\cos t-\tfrac{\sqrt3}2\sin t} 
\\
\dots
\\
\text{(use trigonometry to combine and get rid of radicals)}
\\
\dots,
\end{align} 
and it turns out that
on the given interval
both sides are equivalent to the same expression
\begin{align} 
\sin(\tfrac12t+\tfrac\pi3)
.
\end{align} 

Edit:
\begin{align}
\sqrt{1+\cos\tfrac\pi3\cos t+\sin\tfrac\pi3\sin t} 
&\overset?=
\sqrt{\vphantom{\tfrac{\sqrt3}2}1-(1-2\sin^2\tfrac t2)}+ 
\sqrt{1+\cos\tfrac\pi3\cos t-\sin\tfrac\pi3\sin t}  
,\\
\sqrt{1+\cos(t-\tfrac\pi3)} 
&\overset?=
\sqrt{\vphantom{\tfrac{\sqrt3}2}1-(1-2\sin^2\tfrac t2)}+ 
\sqrt{1+\cos(t+\tfrac\pi3)}
,\\
\sqrt{1+2\cos^2\frac{t-\tfrac\pi3}2-1} 
&\overset?=
\sqrt2\sin\tfrac t2+ 
\sqrt{1+2\cos^2\frac{t+\tfrac\pi3}2-1}
,\\
\cos(\tfrac t2-\tfrac\pi6)
&\overset?=
\sin\tfrac t2+ 
\cos(\tfrac t2+\tfrac\pi6)
,\\
\cos\tfrac t2\cos\tfrac\pi6+
\sin\tfrac t2\sin\tfrac\pi6
&\overset?=
\sin\tfrac t2+ 
\cos\tfrac t2\cos\tfrac\pi6-
\sin\tfrac t2\sin\tfrac\pi6
,\\
\cos\tfrac t2\cos\tfrac\pi6+
\sin\tfrac t2\sin\tfrac\pi6
&\overset?=
\cos\tfrac t2\cos\tfrac\pi6+
\sin\tfrac t2\sin\tfrac\pi6
.
\end{align}
A: Hint:
Square
$$\sqrt{2+\sqrt{3-3b^2}+b}- \sqrt{2-\sqrt{3-3b^2}+b} = \sqrt{2-2b}.$$
This gives
$$2+\sqrt{3-3b^2}+b+2-\sqrt{3-3b^2}+b-2\sqrt{(2+\sqrt{3-3b^2}+b)(2-\sqrt{3-3b^2}+b)}= 2-2b,$$
$$4+2b-2\sqrt{(2+b)^2-(3-3b^2)}= 2-2b,$$
$$2+4b-2\sqrt{4b^2+4b+1}= 0.$$
Remains to discuss the domain and the signs.
A: First, we rearrange the equation into the equivalent equation
$$
\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}} \overset?= \sqrt{2-2b}
$$
which we want to prove. Both sides are positive (keep in mind we have not yet proven the equality), so we can square both sides, and equivalence still holds:
$$
4+2b-2\sqrt{(2+b)^2-(3-3b^2)}
= \left(\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}}\right)^2
\overset?= 2-2b
$$
Ie, $x^2=y^2$ implies $x=y$ if we know $x,y\ge 0$, otherwise the squares could be equal, but the two sides have opposite signs.
Now, simplify the left-hand side,
$$
2-2b = 4+2b-2\sqrt{1+4b+4b^2}= 4+2b-2\sqrt{(2+b)^2-(3-3b^2)} \overset?= 2-2b,
$$
and we see that there actually is equality.
Note that I have used throughout the assumption that the expression inside square roots are non-negative numbers. That's where limitations on $b$ come from.
