Question

Find the value of \[2{\tan ^{ - 1}}x\] if \[x > 1,\].
A) \[{\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\]
B) \[\pi + {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)\]
C) \[\pi + {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\]
D) \[\pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\]

Answer 1

Hint: Substitute \[x = \tan \theta \] and then see what value we are getting and after that substitute the same thing in each and every option to see which option correctly matches.

Complete step-by-step answer:
Substituting \[x = \tan \theta \] in \[2{\tan ^{ - 1}}x\]
Now \[{\tan ^{ - 1}}(\tan \theta ) = \theta \] Therefore \[2{\tan ^{ - 1}}(\tan \theta ) = 2\theta \]
So all we need to see that which one of these options can get me \[2\theta \] after substituting \[x = \tan \theta \]
For option D:
\[\pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\]
After substitution it becomes
\[\pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right)\]
From trigonometric identity we know that \[\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right) = \cos 2\theta \]
Therefore the whole thing become
\[\begin{array}{l}
 = \pi - {\cos ^{ - 1}}\left( {\cos 2\theta } \right)\\
 = \pi - 2\theta
\end{array}\]
Which is clearly the incorrect option.
For Option C:
After substitution it becomes \[\pi + {\tan ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 - {{(\tan \theta )}^2}}}} \right)\]
We know that \[\dfrac{{2\tan \theta }}{{1 - {{(\tan \theta )}^2}}} = \tan 2\theta \]
Therefore it becomes
\[\begin{array}{l}
 = \pi + {\tan ^{ - 1}}\left( {\tan 2\theta } \right)\\
 = \pi + 2\theta
\end{array}\]
Which is an incorrect option.
For option B:
After substitution it becomes \[\pi + {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{(\tan \theta )}^2}}}} \right)\]
We know that \[\left( {\dfrac{{2\tan \theta }}{{1 + {{(\tan \theta )}^2}}}} \right) = \sin 2\theta \]
Therefore the whole thing becomes
\[\begin{array}{l}
 = \pi + {\sin ^{ - 1}}\left( {\sin 2\theta } \right)\\
 = \pi + 2\theta
\end{array}\]
Which again is incorrect
For option A:
After substitution it becomes \[{\cos ^{ - 1}}\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right)\]
As in option D we already know that
\[{\cos ^{ - 1}}\left( {\dfrac{{1 - {{(\tan \theta )}^2}}}{{1 + {{(\tan \theta )}^2}}}} \right) = 2\theta \]
Therefore it is clear that Option A is the correct option here.

Note: We must check each and every option carefully as these types of questions may contain multiple correct options so make sure you check each one of them before writing the final answer.