Question

Find the value of $ { \tan ^{ - 1}}( \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}}) $ .

Answer 1

Hint: Simplify the expression $ \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} $ such that it becomes equal to \[ \tan 2x \].
Use the identities: $ \tan x = \dfrac{{ \sin x}}{{ \cos x}} $ , \[ \sin 2 \theta { \text{ }} = { \text{ }}2 \sin \theta \cos \theta \] , and \[ \cos 2 \theta = { \cos ^2} \theta - { \text{si}}{{ \text{n}}^2} \theta \] for the simplification.
Finally, use this fact: $ { \tan ^{ - 1}}( \tan x) = x $ for \[ \dfrac{{ - \pi }}{2} < x < \dfrac{ \pi }{2} \]to arrive at the answer.

Complete step-by-step answer:
We are given a trigonometric expression $ { \tan ^{ - 1}}( \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}})....(1) $ whose value needs to be computed.
We will first simplify the expression inside the bracket.
So, consider the expression $ \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} $
First, we will use the reciprocal identity: $ \tan x = \dfrac{{ \sin x}}{{ \cos x}} $
Substituting this value of \[{ \text{tan }}x \], we get:
 $
 \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} \ \
   = \dfrac{{2( \dfrac{{ \sin x}}{{ \cos x}})}}{{1 - {{( \dfrac{{ \sin x}}{{ \cos x}})}^2}}} \ \
  $
We know that $ {( \dfrac{{ \sin x}}{{ \cos x}})^2} = \dfrac{{{{ \sin }^2}x}}{{{{ \cos }^2}x}} $ .
Thus, we have
 $
 \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} \ \
   = \dfrac{{ \dfrac{{2 \sin x}}{{ \cos x}}}}{{1 - \dfrac{{{{ \sin }^2}x}}{{{{ \cos }^2}x}}}} \ \
  $
On further simplification of the denominator of this new fraction, we get:
 \[
 \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} \ \
   = \dfrac{{ \dfrac{{2 \sin x}}{{ \cos x}}}}{{ \dfrac{{{{ \cos }^2}x - {{ \sin }^2}x}}{{{{ \cos }^2}x}}}} \ \
  \]
We can eliminate one \[{ \text{cos }}x \] each from the fractions in the numerator and the denominator.
This will leave us with \[2 \sin x \] in the numerator and a fraction in the denominator.
 \[
 \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} \ \
   = \dfrac{{2 \sin x}}{{ \dfrac{{{{ \cos }^2}x - {{ \sin }^2}x}}{{ \cos x}}}} \ \
  \]
We can rewrite the RHS of this equation as follows:
 \[
 \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} \ \
   = \dfrac{{2 \sin x}}{{ \dfrac{{{{ \cos }^2}x - {{ \sin }^2}x}}{{ \cos x}}}} \ \
   = \dfrac{{2 \sin x}}{{ \dfrac{1}{{ \cos x}}({{ \cos }^2}x - {{ \sin }^2}x)}} \ \
  \]
Now, multiply \[{ \text{cos }} x \]in the numerator and the denominator.
 \[
 \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} \ \
   = \dfrac{{2 \sin x \cos x}}{{ \dfrac{{ \cos x}}{{ \cos x}}({{ \cos }^2}x - {{ \sin }^2}x)}} \ \
   = \dfrac{{2 \sin x \cos x}}{{{{ \cos }^2}x - {{ \sin }^2}x}} \ \
  \]
Here, we will use one of the double angle identities in the numerator: For any $ \theta $ , \[{ \text{sin }}2 \theta = 2 \sin \theta \cos \theta \].
Then the expression becomes
 \[ \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} = \dfrac{{ \sin 2x}}{{{{ \cos }^2}x - {{ \sin }^2}x}} \]
Now, let’s use another double angle identity but this time in the denominator.
For any $ \theta $ , \[{ \text{cos }}2 \theta = { \cos ^2} \theta - { \sin ^2} \theta \]
This simplifies the right hand side of our equation even more.
We get
 \[ \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} = \dfrac{{ \sin 2x}}{{ \cos 2x}} \]
We will use the reciprocal identity again: $ \tan x = \dfrac{{ \sin x}}{{ \cos x}} $
 \[ \Rightarrow \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}} = \tan 2x \]
Going back to the original question (1) and substituting this value, we get
 $ \Rightarrow { \tan ^{ - 1}}( \dfrac{{2 \tan x}}{{1 - {{ \tan }^2}x}}) = { \tan ^{ - 1}}( \tan 2x) $
We know that $ \tan \theta $ is not defined at $ - \dfrac{ \pi }{2} $ and $ \dfrac{ \pi }{2} $ .
Also, $ \tan \theta $ is a periodic function with the period $ \pi $ .
Therefore, $ { \tan ^{ - 1}}( \tan \theta ) = \theta $ for $ - \dfrac{ \pi }{2} < \theta < \dfrac{ \pi }{2} $
So, $ \tan 2x $ also has a period of $ \pi $ .
 $ \Rightarrow { \tan ^{ - 1}}( \tan 2x) = 2x $ for \[ - \dfrac{ \pi }{2} < 2x < \dfrac{ \pi }{2} \] or \[ - \dfrac{ \pi }{4} < x < \dfrac{ \pi }{4} \](This is obtained by throughout dividing \[ - \dfrac{ \pi }{2} < 2x < \dfrac{ \pi }{2} \] by 2)
Hence the required answer is \[2x \].

Note: A common mistake that we can often notice among students is that they substitute \[{ \cos ^2}x - { \sin ^2}x = 1 \]. The confusion is mostly because of the similarity it holds with the Pythagorean identity \[{ \cos ^2}x + { \sin ^2}x = 1 \]. However, such a mistake must be avoided at all costs.