Question

The value of ${{\tan }^{-1}}\left[ \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right],\left| x \right|<\dfrac{1}{2},x\ne 0$, is equal to.
(a) $\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}{{x}^{2}}$
(b) $\dfrac{\pi }{4}+{{\cos }^{-1}}{{x}^{2}}$
(c) $\dfrac{\pi }{4}-{{\cos }^{-1}}{{x}^{2}}$

Answer 1

Hint: To solve this question, we will use substitution method. We will take the value of ${{x}^{2}}=\cos \theta $. From trigonometric ratios of half angles, we know that $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$. We will use this relation and reduce it in simples form of trigonometric ratio tan. Once we get the expression in tan, we can use the relation ${{\tan }^{-1}}\left( \tan x \right)=x$. After this, we will reverse the substitution and write the answer in terms of x.

Complete step-by-step solution:
The expression given to us is ${{\tan }^{-1}}\left[ \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]$.
We will substitute ${{x}^{2}}=\cos \theta $. Thus, the expression changes as follows:
$\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }} \right]$
From the trigonometric ratios of half angles we know that $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$.
$\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{2{{\cos }^{2}}\dfrac{\theta }{2}}+\sqrt{2{{\sin }^{2}}\dfrac{\theta }{2}}}{\sqrt{2{{\cos }^{2}}\dfrac{\theta }{2}}-\sqrt{2{{\sin }^{2}}\dfrac{\theta }{2}}} \right]$
We will take $\sqrt{2}$ as common and bring out the sin and cos trigonometric ratio out of the root. The changed expression is as follows:
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{2}\left( \cos \dfrac{\theta }{2}+\sin \dfrac{\theta }{2} \right)}{\sqrt{2}\left( \cos \dfrac{\theta }{2}-\sin \dfrac{\theta }{2} \right)} \right]\]
$\sqrt{2}$ divides out from the numerator and denominator.
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{\theta }{2}+\sin \dfrac{\theta }{2} \right)}{\left( \cos \dfrac{\theta }{2}-\sin \dfrac{\theta }{2} \right)} \right]\]
We will now take $\cos \dfrac{\theta }{2}$ common in the numerator and denominator. From the basic trigonometric ratios, we know that $\dfrac{\sin x}{\cos x}=\tan x$.
Thus, the expression of inverse tan will change as
\[\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{\cos \dfrac{\theta }{2}\left( 1+\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}} \right)}{\cos \dfrac{\theta }{2}\left( 1-\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}} \right)} \right] \\
 & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{\cos \dfrac{\theta }{2}\left( 1+\tan \dfrac{\theta }{2} \right)}{\cos \dfrac{\theta }{2}\left( 1-\tan \dfrac{\theta }{2} \right)} \right] \\
\end{align}\]
The trigonometric ratio of cos divides out from the numerator and the denominator.
$\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\left( 1+\tan \dfrac{\theta }{2} \right)}{\left( 1-\tan \dfrac{\theta }{2} \right)} \right]$
From basic trigonometric ratios, we know that $\tan \dfrac{\pi }{4}=1$. Thus, we will substitute it in the inverse tan expression.
$\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\tan \dfrac{\pi }{4}+\tan \dfrac{\theta }{2}}{\tan \dfrac{\pi }{4}-\tan \dfrac{\theta }{2}} \right]$
Now, we also know that $\dfrac{\tan \dfrac{\pi }{4}+\tan \dfrac{\theta }{2}}{\tan \dfrac{\pi }{4}-\tan \dfrac{\theta }{2}}=\tan \left( \dfrac{\pi }{4}+\dfrac{\theta }{2} \right)$
Therefore, the expression now will be ${{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}+\dfrac{\theta }{2} \right) \right]$,
We know that ${{\tan }^{-1}}\left( \tan x \right)=x$
 $\Rightarrow {{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}+\dfrac{\theta }{2} \right) \right]=\dfrac{\pi }{4}+\dfrac{\theta }{2}$
Now, the substitution we initially made was ${{x}^{2}}=\cos \theta $. Therefore, $\theta ={{\cos }^{-1}}\left( {{x}^{2}} \right)$.
Thus, the expression will be as follows:
$\Rightarrow \dfrac{\pi }{4}+\dfrac{\theta }{2}=\dfrac{\pi }{4}+\dfrac{{{\cos }^{-1}}\left( {{x}^{2}} \right)}{2}$
Therefore, ${{\tan }^{-1}}\left[ \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\dfrac{\pi }{4}+\dfrac{{{\cos }^{-1}}\left( {{x}^{2}} \right)}{2}$.
Hence, option (a) is the correct option.


Note: It is advisable to look at the options before starting to solve the question. They often give us a hint of how to solve the question. In the case of this question, we can see that in every option, one term is ${{\cos }^{-1}}\left( {{x}^{2}} \right)$, so, we can deduce that we need to make a substitution of ${{x}^{2}}=\cos \theta $, which will be reversed at the end of the solution.