Prove the given inverse trigonometric expression expression, \[{{\tan }^{-1}}\sqrt{x}=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right),x\in (0,1)\] .

Hint: In this question, we have different inverse trigonometric functions in LHS and RHS. In LHS, we have tan inverse function while in RHS, we have cos inverse function. So, we have to transform the inverse tan function into an inverse cosine function. We consider \[\theta ={{\tan }^{-1}}\sqrt{x}\]. Then transform \[\tan \theta \] into \[\cos \theta \] . Use formula, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] and solve it further.

Complete step-by-step solution -
Solving LHS part, let us assume, \[\theta ={{\tan }^{-1}}\sqrt{x}\]……………(1)
Taking tan in both LHS as well as RHS in equation (1), we get
\[\tan \theta =\sqrt{x}\]………………..(2)
As in the RHS, we don’t have any \[\sqrt{x}\]. So, we have to remove \[\sqrt{x}\].
Now, squaring LHS and RHS of equation(2), we get
\[\tan \theta =\sqrt{x}\]
\[\Rightarrow {{\tan }^{2}}\theta =x\]……………….(3)
As the RHS has inverse cosine function, so we have to switch tan function into cosine function. To do so, we have a formula,
\[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]
Using equation(3), we can write it as
\[{{\sec }^{2}}\theta -x=1\]
\[\Rightarrow {{\sec }^{2}}\theta =1+x\]………………..(4)
As the RHS has inverse cosine function, we have to convert the above equation in terms of cosine. We know the relation between sec and cos, that is
\[\sec \theta =\dfrac{1}{\cos \theta }\]…………(5)
Putting equation(5) in equation(4), we get
\[\begin{align}
  & {{\sec }^{2}}\theta =1+x \\
 & \Rightarrow \dfrac{1}{{{\cos }^{2}}\theta }=1+x \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{1+x} \\
\end{align}\]
We know the identity, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] .
Putting, \[{{\cos }^{2}}\theta =\dfrac{1}{1+x}\] in the above equation, we get
\[\begin{align}
  & \cos 2\theta =2.\left( \dfrac{1}{1+x} \right)-1 \\
 & \Rightarrow \cos 2\theta =\dfrac{2}{1+x}-1 \\
 & \Rightarrow \cos 2\theta =\dfrac{2-1-x}{1+x} \\
 & \Rightarrow \cos 2\theta =\dfrac{1-x}{1+x} \\
\end{align}\]
We can also write the above expression as,
\[\begin{align}
  & \cos 2\theta =\dfrac{1-x}{1+x} \\
 & \Rightarrow 2\theta ={{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right) \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\]………………………………(6)
Equation(1) and equation(6) are equal.
\[\theta ={{\tan }^{-1}}\sqrt{x}=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\] .
Therefore, LHS=RHS.
Hence, proved.

Note: This question can also be solved by using the Pythagoras theorem.
Assume,
\[\begin{align}
  & \theta ={{\tan }^{-1}}\sqrt{x} \\
 & \Rightarrow \tan \theta =\sqrt{x} \\
\end{align}\] ------(7)
Here, we know that $\tan \theta = \dfrac{\text{Base}}{\text{Height}} $ ----(8)
In the comparing equation(7) and equation(8) we get,
 Base = $\sqrt{x}$ and Height = 1.
Now, using a right-angled triangle, we can get the value of \[\cos \theta \].

Using Pythagoras theorem, we can find the hypotenuse.
Hypotenuse = \[\sqrt{{{\left( height \right)}^{2}}+{{\left( base \right)}^{2}}}\]
\[\begin{align}
  & \sqrt{{{\left( \sqrt{x} \right)}^{2}}+{{1}^{2}}} \\
 & =\sqrt{x+1} \\
\end{align}\]
\[\begin{align}
  & \cos \theta =\dfrac{\text{base}}{\text{hypotenuse}} \\
 & \cos \theta =\dfrac{1}{\sqrt{x+1}} \\
\end{align}\]
Now, using the identity, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] , we can solve it further.