If  -1 <  x < 0 , then  sin^-1x equals

$\tan^{-1}\frac{x}{\sqrt{1-x^2}}$

Since -1< x < 0, -π/2 < sin^–1x < 0.

Let sin^-1 x = α i.e. sinα = x

Then $tanα=\frac{x}{\sqrt{1-x^2}}⇒ α = tan^{-1}\frac{x}{\sqrt{1-x^2}}⇒ sin^{-1}x = tan^{-1}\frac{x}{\sqrt{1-x^2}}$.

Hence (B) is the correct answer.

Alternative:

$sin^{-1}x=-sin^{-1}(-x)=-tan^{-1}(-\frac{x}{\sqrt{1-x^2}})=tan^{-1}\frac{x}{\sqrt{1-x^2}}$