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.