2014/078) given: $(1+\sqrt{1+x}\tan\ x=1+\sqrt{1-x}$ then prove that $\sin\ 4x=x$

Let $x = \sin\ 2y$

Then $\sqrt{1+x} = \sin\ y + \cos\ y$

and $\sqrt{1-x} = \sin\ y - \cos\ y$

hence

$\tan\ x = \dfrac{1+ \sin\ y - \cos\ y}{1+ \sin\ y + \cos\ y}$

using if $\dfrac{a}{b} =\dfrac{c}{d}$ => $\dfrac{a+b}{a-b} =\dfrac{c+d}{c-d}$

$\dfrac{\tan\ x + 1}{\tan\ x -1} = \dfrac{ 1+ \sin\ y}{-\cos\ y}$

$\dfrac{\sin\ x + \cos\ x }{\sin\ x -\cos\ x} = \dfrac{ 1+ \sin\ y}{-\cos\ y}$

$\dfrac{(\sin\ x + \cos\ x) ^2}{\sin ^2 x -\cos ^2 x} = \dfrac{1+ \sin\ y}{ \cos\ y}$

$\dfrac{1+ \sin\ 2x}{-\cos\ 2x} = \dfrac{ 1+\ sin\ y}{ \cos\ y}$

From the above $y = 2x$ so $x = \sin\ 4x$ proved