If $ -1 < x < 0,$ then $sin^{-1}x $ equals

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

Let $sin^{-1}x = \theta $. Then , x = sin $\theta $.

Now, $-1 < x < 0 ⇒ -\frac{\pi}{2} < \theta < 0 $

$\pi - sin^{-1}\sqrt{1-x^2}$

$= \pi - sin^{-1} (cos \theta )$

$= \pi - sin (\frac{\pi}{2}+\theta ) = \pi - (\frac{\pi}{2}+\theta )=\frac{\pi}{2}+\theta $

As - $\frac{\pi}{2} < \theta < 0$

$∴ 0< \pi - \theta <\frac{\pi}{2} ⇒ 0 < \pi 0 sin^{-1}\sqrt{1-x^2} < \frac{\pi}{2}$

$∴ sin^{-1} x ≠ \pi - sin^{_1} \sqrt{1-x^2}$

So, option (a) is not correct.

We have,

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

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

Thus, option (b) is correct.