Inverse Trigonometric Functions-basic

given equation:

$sin^{-1}x+sin^{-1}(1-x)=cos^{-1}x$

or, $1-x=sin(cos^{-1}x-sin^{-1}x)$

or, $1-x=sin(cos^{-1}x ). cos(sin^{-1}x) -cos(cos^{-1}x) . sin(sin^{-1}x)$

or, $1-x=(sin .sin^{-1} \sqrt{1-x^2} ).(cos . cos^{-1} \sqrt{1-x^2})-x^2$

or, $1-x=\sqrt{1-x^2} .\sqrt{1-x^2} -x^2$

or, $1-x=1-x^2-x^2$

or, $2x^2-x=0$

or, $x(2x-1)=0$

so now,

$x=0$ , or $x=\frac{1}{2}$

Because it multiple choice : only the last solution is possibly valid. The others give at least 1 input outside of the domains of the function. Testing last value gives the equation holds for X = 0 and x = 0.5