Question

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

Answer 1

Hint: We will use the substitution $x={{\tan }^{2}}\left( \theta \right)$ here to solve the question further. Using this substitution we will convert the right side of the expression ${{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right]$ in terms of inverse tan.

Complete step-by-step answer:
Now we will first consider the expression ${{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right]....(i)$.
We will substitute $x={{\tan }^{2}}\left( \theta \right)$ here to the right side of the equation (i). Therefore, we have
$\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\left( \theta \right)}{1+{{\tan }^{2}}\left( \theta \right)} \right)\,$. As we know that $\tan \left( p \right)=\dfrac{\sin \left( p \right)}{\cos \left( p \right)}$. Therefore we will have
$\begin{align}
  & \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-\dfrac{{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}}{1+\dfrac{{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}} \right)\, \\
 & \Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{\dfrac{{{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}}{\dfrac{{{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}} \right)\, \\
\end{align}$
After cancelling the term ${{\cos }^{2}}\left( \theta \right)$ we will have $\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)} \right)$. As we know that ${{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)=1$. Therefore, we will get
$\begin{align}
  & \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)}{1} \right) \\
 & \Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( {{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right) \right) \\
\end{align}$
Now we will apply the formula given by $\cos \left( 2\theta \right)={{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)$ in the above expression thus, we will get
$\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \cos \left( 2\theta \right) \right)$
As we know that ${{\cos }^{-1}}\left( \cos \left( x \right) \right)=x$ therefore, we have
$\begin{align}
  & \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}\times 2\theta \\
 & \Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\theta ...(ii) \\
\end{align}$
Since, we have that $x={{\tan }^{2}}\left( \theta \right)$. By taking root to both the sides of the term $x={{\tan }^{2}}\left( \theta \right)$ we will get $\sqrt{x}=\tan \left( \theta \right)$. Now we will take the tan term to the left side of the expression we will get ${{\tan }^{-1}}\left( \sqrt{x} \right)=\theta $. Now we will substitute the value of $\theta $ in equation (ii). Therefore, we have
$\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,={{\tan }^{-1}}\left( \sqrt{x} \right)$ which is our required expression.
Hence, the expression ${{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right]$ is proved.

Note: Alternatively we could have used the method in which we can convert ${{\cos }^{-1}}$ term into ${{\tan }^{-1}}$. This can be done by the formula ${{\cos }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$. We can use this formula here since we have the condition $x\in \left[ 0,1 \right]$ which is the required condition for the formula.