Answer
$x = 0$
Work Step by Step
We can use the following identity:
$sin^{-1}~a+sin^{-1}~b = sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$
We can find the solution:
$sin^{-1}~x+tan^{-1}~x = 0$
$sin^{-1}~x+sin^{-1}~\frac{x}{\sqrt{x^2+1}} = 0$
$sin^{-1}~(x~\sqrt{1-\frac{x^2}{x^2+1}}+\frac{x}{\sqrt{x^2+1}}\sqrt{1-x^2}) = 0$
$x~\sqrt{1-\frac{x^2}{x^2+1}}+\frac{x}{\sqrt{x^2+1}}\sqrt{1-x^2} = 0$
$\frac{x}{\sqrt{x^2+1}}+\frac{x\sqrt{1-x^2} }{\sqrt{x^2+1}}= 0$
$x+x\sqrt{1-x^2} = 0$
$x(1+\sqrt{1-x^2}) = 0$
$x = 0$
