Question

Find the value of ${\tan ^{ - 1}}(\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}}) - {\tan ^{ - 1}}(\dfrac{{2x}}{{1 - {x^2}}})$.

Answer 1

Hint: To solve this equation, you have to find out what's the formula for ${\tan ^{ - 1}}3x$ and ${\tan ^{ - 1}}2x$ . So, now you know that formula so compare it to given equation and convert given equation to that formula after that you have to apply formula for subtraction in ${\tan ^{ - 1}}x$ and ${\tan ^{ - 1}}y$ after that you have to simple mathematics calculation to find the answer.

Complete step by step answer:
On the very first time, we have to write our equation,
$ \Rightarrow {\tan ^{ - 1}}(\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}}) - {\tan ^{ - 1}}(\dfrac{{2x}}{{1 - {x^2}}})$
Now, let’s see the formula for ${\tan ^{ - 1}}3x$ and ${\tan ^{ - 1}}2x$ ,
$ \Rightarrow {\tan ^{ - 1}}3x = {\tan ^{ - 1}}(\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}})$
$ \Rightarrow {\tan ^{ - 1}}2x = {\tan ^{ - 1}}(\dfrac{{2x}}{{1 - {x^2}}})$
You can clearly see that given equation is nothing but formula for ${\tan ^{ - 1}}3x$ and ${\tan ^{ - 1}}2x$, so rewrite our problem and we will get,
$ \Rightarrow \;{\tan ^{ - 1}}3x - {\tan ^{ - 1}}2x$
Now, let’s see formula for ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y$ ,
$ \Rightarrow {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}(\dfrac{{x - y}}{{1 + xy}})$
Now, apply above formula in our last equation and we will get,
Replacing $x = 3x$ and $y = 2x$ ,
After putting values of $x$ and $y$ we will get,
$ \Rightarrow {\tan ^{ - 1}}(\dfrac{{3x - 2x}}{{1 + (3x)(2x)}})$
After further simplification we will get,
$ \Rightarrow {\tan ^{ - 1}}(\dfrac{x}{{1 + 6{x^2}}})$
This is the required answer.

Additional information:
There are also formula for ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y$ and that is,
$ \Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}(\dfrac{{x + y}}{{1 - xy}})$
But there are also conditions for applying both the formulas. Let’s see both conditions one by one.
Condition for ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y$ :
So, condition for ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y$ is, product of $x$ and $y$ is must have to be less than 1,
$x \times y < 1$
Condition for ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y$ :
So, condition for ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y$ is, product of $x$ and $y$ is must have to be greater than -1,
$x \times y > - 1$

Note:
For this problem, there are no options available so there are two possible answers. First one is we derived and for other answer you have to recall formula for ${\tan ^{ - 1}}3x$,
Formula for ${\tan ^{ - 1}}3x = {\tan ^{ - 1}}2x - {\tan ^{ - 1}}x$
So converting a given problem in the above equation we will get the second answer is ${\tan ^{ - 1}}x$.