If 0 < x < 1, then $tan^{-1}\left(\frac{\sqrt{1-x^2}}{1+x}\right)$ is equal to 

all the above

Let $x = cos \theta $. Then,

$0 < x< 1 ⇒ 0 < cos \theta < 1 ⇒ 0 < \theta <\frac{\pi}{2}$

Now,

$tan^{-1}\left(\frac{\sqrt{1-x^2}}{1+x}\right)$

$= tan^{-1}\sqrt{\frac{1-x}{1+x}}$

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

Thus, option (a) is true

$= cos^{-1}\sqrt{\frac{1+x}{2}}=cos^{-1}\left(cos\frac{\theta}{2}\right)=\frac{1}{2}\theta =\frac{1}{2}cos^{-1}x$

So, options (b) is true.

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

So, options (c) is true.