Solve $\tan ^{ - 1}x + \tan ^{ - 1}2x = n\pi - \tan ^{ - 1}3x$, $n \in I, x \in R$, for n and x

(This question has multiple correct options)

(A). x = 0,n = 0

(B). x = 1,n = 1

(C). x = - 1,n = - 1

(D). x = 2,n = 2

Hint: Let’s split the equation into different cases like when n $\in {{\text{I}}^ + }{\text{ and n}} \in {{\text{I}}^ - }$. Then finding out the value of the equation for different cases will make the problem easy. Now, apply this logic and find out the answer.

Complete step by step answer:

It is given that,

$\Rightarrow {\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(2x) = n\pi - {\tan ^{ - 1}}(3x){\text{ (1)}}$

As we know by trigonometry formula that,

$\Rightarrow {\tan ^{ - 1}}(A) + {\tan ^{ - 1}}(B) = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right){\text{ (2)}}$

So, according to the formula at equation 2 we can write,

$\Rightarrow {\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(2x) = {\tan ^{ - 1}}\left( {\dfrac{{x + 2x}}{{1 - x*2x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{3x}}{{1 - 2{x^2}}}} \right){\text{ (3)}}$

As from equation 1 and equation 3 we see that it can be written that,

$\Rightarrow n\pi - {\tan ^{ - 1}}(3x){\text{ }} = {\tan ^{ - 1}}\left( {\dfrac{{3x}}{{1 - 2{x^2}}}} \right){\text{ (4)}}$

Now, taking tan both sides of the equation 4 we will get,

$\Rightarrow \dfrac{{3x}}{{1 - 2{x^2}}} = \tan \left( {n\pi - {{\tan }^{ - 1}}(3x)} \right){\text{ (5)}}$

Now as we can see from the question that $n \in I,$

So, there can be two cases and that were,

Case 1: - When n $\in {I^ + }$ (when all n as an positive integer)

In this case of equation 5 can be written as,

$\Rightarrow \dfrac{{3x}}{{1 - 2{x^2}}} = - \tan \left( {{{\tan }^{ - 1}}(3x)} \right)$ because $tan\left( {n\pi - \theta } \right){\text{ = - tan}}\left( \theta \right){\text{ when }}n \in {I^ + }{\text{ }}$

$\Rightarrow {\text{So, }}\dfrac{{3x}}{{1 - 2{x^2}}} = - 3x$ as we know that,

As, we know that $\tan \left( {{{\tan }^{ - 1}}(A)} \right) = A$

$\Rightarrow \dfrac{{3x}}{{1 - 2{x^2}}} = - 3x{\text{ (6)}}$

Adding 3x to both sides of equation 6 we get,

$\Rightarrow {\text{3x}}\left( {\dfrac{1}{{1 - 2{x^2}}} + 1} \right) = 0$

On taking LCM of the above equation we get,

$\Rightarrow {\text{ 3x}}\left( {\dfrac{{2 - 2{x^2}}}{{1 - 2{x^2}}}} \right) = 6{\text{x}}\left( {\dfrac{{1 - {x^2}}}{{1 - 2{x^2}}}} \right) = 0{\text{ (7)}}$

Hence from equation 7 we get x = 0 and ${x^2} - 1 = 0$

So, x = 0, $\pm 1$

$\Rightarrow$ Hence for all n $\in \{ 0,1,2....N\}$ we get x = - 1,0,1

Case 2: - When n $\in {I^ - }$ (For all n as negative integer)

In this case of equation 5 can be written as ,

$\dfrac{{3x}}{{1 - 2{x^2}}} = - \tan \left( {n\pi + {{\tan }^{ - 1}}(3x)} \right){\text{ = }} - 3x$ because n is negative and we

know that,

$\Rightarrow {\text{ - tan}}\left( {n\pi + \theta } \right){\text{ = - tan(}}\theta ){\text{ when }}n \in {I^ - }{\text{ here }}\theta = {\tan ^{ - 1}}(3x)$

So, $\dfrac{{3x}}{{1 - 2{x^2}}} = - 3x$ as we know that,

As, $\tan \left( {{{\tan }^{ - 1}}(A)} \right) = A$

$\Rightarrow \dfrac{{3x}}{{1 - 2{x^2}}} = - 3x{\text{ (8)}}$

Now adding 3x to both sides of the equation 8 we get,

$\Rightarrow {\text{ 3x}}\left( {\dfrac{1}{{1 - 2{x^2}}} + 1} \right) = 0$

On taking LCM of the above equation we get,

$\Rightarrow {\text{ 3x}}\left( {\dfrac{{2 - 2{x^2}}}{{1 - 2{x^2}}}} \right) = 6{\text{x}}\left( {\dfrac{{1 - {x^2}}}{{1 - 2{x^2}}}} \right) = 0{\text{ (9)}}$

Hence from equation 9 we get x = 0 and ${x^2} - 1 = 0$

So, x = 0, $\pm 1$

$\Rightarrow$ Hence for all n $\in \{ 0,1,2....N\} {\text{ we get }}x = \{ - 1,0,1\}$

So, from cases 1 and 2 we found that Correct options for the above question are B, C and D.

NOTE: Remember the formula ${\tan ^{ - 1}}(A) + {\tan ^{ - 1}}(B)$ and substitute in values properly in the given conditions. If you make mistakes while substituting the values, there are chances of missing out other right answers.