Question

Write the values of the $x$ for which $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}$ , holds.

Answer 1

Hint : In this question, we will find the value of the x for which $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}$ , holds.
For that, we will consider ${\tan ^{ - 1}}x = y$ and determine $x$ . And then, substitute the $x$ in the RHS of the given and evaluate it and by knowing the given think of which trigonometric identity is suitable and apply it to evaluate the given. And finally prove LHS=RHS to determine the value of $x$ . Having a good knowledge of inverse trigonometric formulas helps to solve this problem.

Complete step-by-step answer :
Now, we have given that
$2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}$
We need to find the values of $x$ .
Let us consider, ${\tan ^{ - 1}}x = y$
Therefore, $x = \tan y$
Now, substitute $x = \tan y$ in the RHS of the given equation,
RHS $= {\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}$
Therefore, RHS $= {\cos ^{ - 1}}\dfrac{{1 - {{\tan }^2}y}}{{1 + {{\tan }^2}y}}$
As, we know that,
$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
Therefore, by substituting the value, we have,
RHS= ${\cos ^{ - 1}}\left( {\cos 2y} \right)$
Here, cancel the ${\cos ^{ - 1}}$ and $\cos$ . As a result, $\theta$ remains. Hence, we have,
RHS $= 2y$
As we have considered ${\tan ^{ - 1}}x = y$ , substituting the value of $y$ ,
RHS $= 2{\tan ^{ - 1}}x$
Therefore, LHS=RHS
Hence, $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}$ is the identity one can remember for future reference and can remember it as an standard identity for all the values of $x\ge0$ since x for negative values ${\tan ^{ - 1}}(-x)$ = $-{\tan ^{ - 1}}x$.
Therefore $x\ge0$ is required.
So, the correct answer is “ $x\ge0$ ”.

Note : In this question, it is important to note that whenever these types of questions are given, be clear and confident about the identities which helps in the simplification process. However, as same as the process we did above, we can take $x = \tan \theta$ and substitute at both the sides of the equation i.e., in both LHS and RHS. Evaluate both LHS and RHS to the simplest form which gives the required solution. By
knowing the given, we can identify, $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}$ is an identity of inverse trigonometry. Also, $2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$ and $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}}$ . The inverse trigonometric functions are also known as the anti trigonometric functions or sometimes called arcus functions or cyclometric functions.