Question

Solve for x: $ {\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x $

Answer 1

Hint: Here we will make two pairs of two terms from the given expression and apply the sum and difference of two tangent inverse formulas and then accordingly simplify for the resultant answer.

Complete step-by-step answer:
First of all take the given equation –
 $ {\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x $
Move one term from the left hand side of the equation to the right hand side. Remember when you move any term from one side to another, sign also changes. Positive term become negative and vice-versa.
So, above equation can be re-written as –
 $ {\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x - {\tan ^{ - 1}}x $ .... (I)
Take LHS and RHS one by one.
LHS $ = {\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}(x + 1) $
Apply Addition of two inverse trigonometric tangent formulas –
We know that $ {\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right) $
LHS $ = {\tan ^{ - 1}}\left( {\dfrac{{x - 1 + x + 1}}{{1 - (x - 1)(x + 1)}}} \right) $
Simplify the above equation. Like terms with equal values and opposite signs cancel each other. When you open the bracket, the negative sign outside the brackets changes sign of all the terms inside the bracket.
LHS $ = {\tan ^{ - 1}}\left( {\dfrac{{x + x}}{{1 - ({x^2} - 1)}}} \right) $
LHS $ = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2} + 1)}}} \right) $
Simplify the denominator of the above equation –
LHS $ = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{2 - {x^2}}}} \right) $ ..... (A)
Similarly take the RHS of the equation (I)
RHS $ = {\tan ^{ - 1}}3x - {\tan ^{ - 1}}x $
Here apply the difference of two tangent inverse formulas-
 $ {\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right) $
RHS $ = {\tan ^{ - 1}}\left( {\dfrac{{3x - x}}{{1 + (3x)(x)}}} \right) $
When the brackets are opened and there is a positive sign outside the bracket then there is no change in the sign of the terms in the bracket. Simplify the above equation –
RHS $ = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + 3{x^2}}}} \right) $ ..... (B)
Since, LHS $ = $ RHS
Take Right hand side of both sides equal.
 $ {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{2 - {x^2}}}} \right) $ $ = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + 3{x^2}}}} \right) $
Tangent inverse from both the sides of the equation cancel each other.
\[ \Rightarrow \left( {\dfrac{{2x}}{{2 - {x^2}}}} \right) = \left( {\dfrac{{2x}}{{1 + 3{x^2}}}} \right)\]
Do cross-multiplication –
\[ \Rightarrow 2x(1 + 3{x^2}) = 2x(2 - {x^2})\]
Common multiple from both the sides of the equation cancels each other.
\[ \Rightarrow (1 + 3{x^2}) = (2 - {x^2})\]
Take all the constants on one side and variables on other side
\[ \Rightarrow 3{x^2} + {x^2} = 2 - 1\]
Simplify among the like terms –
 $ \Rightarrow 4{x^2} = 1 $
When the term multiplicative on one side moves from one side then it goes to the denominator.
 $ \Rightarrow {x^2} = \dfrac{1}{4} $
Take square-root on both the sides of the equation.
 $ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{1}{4}} $
Square and square-root cancel each other on the left hand side of the equation.
 $ \Rightarrow x = \pm \dfrac{1}{2} $ is the required answer.
So, the correct answer is “$\dfrac{1}{2} $”.

Note: Be careful while changing the terms from one side to another and while opening the brackets. Be very careful about the signs against the terms, it is the most important for the accurate and an efficient solution. Also, refer to the algebraic expressions and formulas.