Prove the given inverse trigonometric expression ${{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}{{\cos }^{-1}}x,\dfrac{-1}{\sqrt{2}}\le x\le 1$.

Hint: Substitute $x=\cos \theta $ and then use the half angle formula given by: $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$. Cancel the common terms and remove the square root by using mod. Now, check whether the expression inside the mod is positive or negative. If it is positive then, remove the mod simply and if it is negative then remove the mod by adding a negative sign in the expression. Now, divide the numerator and denominator by $\cos \dfrac{\theta }{2}$ and use the formula: $\dfrac{1-\tan \theta }{1+\tan \theta }=\tan \left( \dfrac{\pi }{4}-\theta \right)$. Finally, use the identity: ${{\tan }^{-1}}\left( \tan \theta \right)=\theta $ where $ \theta \in(\dfrac{-\pi}{2},\dfrac{\pi}{2}) $ to get the answer.

Complete step-by-step solution -
We have to prove: ${{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}{{\cos }^{-1}}x$
$L.H.S={{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]$
Substituting, $x=\cos \theta $, we get,
$\begin{align}
  & L.H.S={{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }} \right] \\
\end{align}$
Using the formula: $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$, we get,
$L.H.S={{\tan }^{-1}}\left[ \dfrac{\sqrt{2{{\cos }^{2}}\dfrac{\theta }{2}}-\sqrt{2{{\sin }^{2}}\dfrac{\theta }{2}}}{\sqrt{2{{\cos }^{2}}\dfrac{\theta }{2}}+\sqrt{2{{\sin }^{2}}\dfrac{\theta }{2}}} \right]$
Cancelling $\sqrt{2}$ from numerator and denominator, we get,
\[\begin{align}
  & L.H.S={{\tan }^{-1}}\left[ \dfrac{\sqrt{{{\cos }^{2}}\dfrac{\theta }{2}}-\sqrt{{{\sin }^{2}}\dfrac{\theta }{2}}}{\sqrt{{{\cos }^{2}}\dfrac{\theta }{2}}+\sqrt{{{\sin }^{2}}\dfrac{\theta }{2}}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{\left| \cos \dfrac{\theta }{2} \right|-\left| \sin \dfrac{\theta }{2} \right|}{\left| \cos \dfrac{\theta }{2} \right|+\left| \sin \dfrac{\theta }{2} \right|} \right] \\
\end{align}\]
It is given that:
$\begin{align}
  & \dfrac{-1}{\sqrt{2}}\le x\le 1 \\
 & \Rightarrow \dfrac{-1}{\sqrt{2}}\le \cos \theta \le 1 \\
 & \Rightarrow \cos \dfrac{3\pi }{4}\le \cos \theta \le \cos 0 \\
\end{align}$
Since, cosine is decreasing function, therefore, when we will remove cosine function from the expression then sign of the inequality gets reversed.
$\begin{align}
  & \Rightarrow 0\le \theta \le \dfrac{3\pi }{4} \\
 & \Rightarrow 0\le \dfrac{\theta }{2}\le \dfrac{3\pi }{8} \\
\end{align}$
Clearly, $\theta $ lies in the first quadrant and hence, both sine and cosine function will be positive. Therefore, the expression inside the mod is positive. Therefore, L.H.S becomes
\[L.H.S={{\tan }^{-1}}\left[ \dfrac{\cos \dfrac{\theta }{2}-\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}+\sin \dfrac{\theta }{2}} \right]\]
Dividing both numerator and denominator by $\cos \dfrac{\theta }{2}$, we get,
\[\begin{align}
  & L.H.S={{\tan }^{-1}}\left[ \dfrac{1-\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}}{1+\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{1-\tan \dfrac{\theta }{2}}{1+\tan \dfrac{\theta }{2}} \right] \\
\end{align}\]
Using the formula: $\dfrac{1-\tan \theta }{1+\tan \theta }=\tan \left( \dfrac{\pi }{4}-\theta \right)$, we get,
\[L.H.S={{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}-\dfrac{\theta }{2} \right) \right]\]
Now, using the identity: ${{\tan }^{-1}}\left( \tan \theta \right)=\theta $, where $\theta \in (\dfrac{-\pi}{2},\dfrac{\pi}{2})$ we get,
\[L.H.S=\left( \dfrac{\pi }{4}-\dfrac{\theta }{2} \right)\]
Since we have assumed, $x=\cos \theta $, therefore, $\theta ={{\cos }^{-1}}x$.
Substituting, $\theta ={{\cos }^{-1}}x$, we get,
\[\begin{align}
  & L.H.S=\left( \dfrac{\pi }{4}-\dfrac{{{\cos }^{-1}}x}{2} \right) \\
 & =R.H.S \\
\end{align}\].

Note: One may note that while removing the mod, we have to be careful about the conditions given, that is, the range of ‘x’. Positive or negative value of a trigonometric function depends on the quadrant in which the angle is lying. In the above question, the angle $\dfrac{\theta }{2}$ lies in the first quadrant, therefore, the value of the trigonometric function inside the mod is positive.