Question & Answer
QUESTION

Solution of the equation \[{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}x+{{\tan }^{^{-1}}}(x+1)={{\tan }^{^{-1}}}3x\] is:
A.x = 0
B.$x=\pm \dfrac{1}{2}$
C.$x=\pm \dfrac{1}{3}$
D.None of this.

ANSWER Verified Verified
Hint: Shift the term ‘\[{{\tan }^{-1}}x\]’ on the right hand side of the equation and the use the formulae \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\] and \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left[ \dfrac{x-y}{1+xy} \right]\] to simplify the equation to get the value of ‘x’.

Complete step-by-step answer:
To solve the given problem we will write the given equation first, therefore,
\[{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}x+{{\tan }^{^{-1}}}(x+1)={{\tan }^{^{-1}}}3x\]
By rearranging the above equation we will get,
\[{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}(x+1)+{{\tan }^{^{-1}}}x={{\tan }^{^{-1}}}3x\]
To proceed further in the solution we should know the formula given below,
Formula:
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]
If we use the above formula for the first two terms of the given equation we will get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\left( x-1 \right)+\left( x+1 \right)}{1-\left( x-1 \right)\left( x+1 \right)} \right]+{{\tan }^{^{-1}}}x={{\tan }^{^{-1}}}3x\]
By simplifying the above equation we will get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{x-1+x+1}{1-\left( x-1 \right)\left( x+1 \right)} \right]+{{\tan }^{^{-1}}}x={{\tan }^{^{-1}}}3x\]
To proceed further in the solution we should know the formula given below,
Formula:
\[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
If we use the above formula in the given equation we will get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{2x}{1-\left( {{x}^{2}}-{{1}^{2}} \right)} \right]+{{\tan }^{^{-1}}}x={{\tan }^{^{-1}}}3x\]
Further simplification in the above equation will give,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{2x}{1-{{x}^{2}}+1} \right]+{{\tan }^{^{-1}}}x={{\tan }^{^{-1}}}3x\]
If we shift the second term of the above equation on the right hand side of the equation we will get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right)={{\tan }^{^{-1}}}3x-{{\tan }^{^{-1}}}x\] ……………………………………….. (1)
Now before we solve further we should know the formula given below,
Formula:
\[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left[ \dfrac{x-y}{1+xy} \right]\]
If we use the above formula on the right hand side of the equation we will get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right)={{\tan }^{-1}}\left[ \dfrac{3x-x}{1+\left( 3x \right)\times \left( x \right)} \right]\]
If we simplify the above equation we will get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right)={{\tan }^{-1}}\left[ \dfrac{2x}{1+3{{x}^{2}}} \right]\]
Taking ‘tan’ on the both sides of the equation will give,
\[\Rightarrow \tan \left[ {{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right) \right]=\tan \left[ {{\tan }^{-1}}\left[ \dfrac{2x}{1+3{{x}^{2}}} \right] \right]\] ……………………………….. (2)
Now to proceed further in the solution we should know the formula given below,
Formula:
\[\tan \left( {{\tan }^{-1}}x \right)=x\]
If we use the above formula in equation (2) we will get,
\[\Rightarrow \dfrac{2x}{2-{{x}^{2}}}=\dfrac{2x}{1+3{{x}^{2}}}\]
By doing cross multiplication in the above equation we will get,
\[\Rightarrow 2x\left( 1+3{{x}^{2}} \right)=2x\left( 2-{{x}^{2}} \right)\]
If we shift the ‘2x’ on the right had side of the equation we will get,
\[\Rightarrow \left( 1+3{{x}^{2}} \right)=\dfrac{2x\left( 2-{{x}^{2}} \right)}{2x}\]
By simplifying the above equation we will get,
\[\Rightarrow 1+3{{x}^{2}}=2-{{x}^{2}}\]
If we shift the ‘\[-{{x}^{2}}\]’ on the left hand side and 1 on the right hand side of the equation we will get,
\[\Rightarrow {{x}^{2}}+3{{x}^{2}}=2-1\]
If we simplify the above equation we will get,
\[\Rightarrow 4{{x}^{2}}=1\]
Further simplification in the above equation will give,
\[\Rightarrow {{x}^{2}}=\dfrac{1}{4}\]
Taking square roots on both sides of the above equation we will get,
\[\Rightarrow x=\pm \dfrac{1}{2}\]

Therefore the solution of the equation \[{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}x+{{\tan }^{^{-1}}}(x+1)={{\tan }^{^{-1}}}3x\] is \[x=\pm \dfrac{1}{2}\].
Therefore the correct answer is option (b).

Note: At the step \[\tan \left[ {{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right) \right]=\tan \left[ {{\tan }^{-1}}\left[ \dfrac{2x}{1+3{{x}^{2}}} \right] \right]\] you can directly cancel \[{{\tan }^{-1}}\] from both sides of the equation without taking ‘tan’ on both sides if you are solving it in competitive exam. But in a descriptive answer if you ignore this step then there may be marked reduction.