Solveeit Logo
Login

Question

Question: Solution of the equation \[{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}x+{{\tan }^{^{-1}}}(x+1)={{\tan }...

Solution of the equation tan⁡−1(x−1)+tan⁡−1x+tan⁡−1(x+1)=tan⁡−13x{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}x+{{\tan }^{^{-1}}}(x+1)={{\tan }^{^{-1}}}3x is:
A.x = 0
B.x=±12x=\pm \dfrac{1}{2}
C.x=±13x=\pm \dfrac{1}{3}
D.None of this.

Explanation

Solution

Hint: Shift the term ‘tan⁡−1x{{\tan }^{-1}}x’ on the right hand side of the equation and the use the formulae tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) and tan⁡−1x−tan⁡−1y=tan⁡−1[x−y1+xy]{{\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⁡−1x+tan⁡−1(x+1)=tan⁡−13x{{\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⁡−1x=tan⁡−13x{{\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⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\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,
⇒tan⁡−1[(x−1)+(x+1)1−(x−1)(x+1)]+tan⁡−1x=tan⁡−13x\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,
⇒tan⁡−1[x−1+x+11−(x−1)(x+1)]+tan⁡−1x=tan⁡−13x\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:
(a+b)(a−b)=a2−b2\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,
⇒tan⁡−1[2x1−(x2−12)]+tan⁡−1x=tan⁡−13x\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,
⇒tan⁡−1[2x1−x2+1]+tan⁡−1x=tan⁡−13x\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,
⇒tan⁡−1(2x2−x2)=tan⁡−13x−tan⁡−1x\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⁡−1x−tan⁡−1y=tan⁡−1[x−y1+xy]{{\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,
⇒tan⁡−1(2x2−x2)=tan⁡−1[3x−x1+(3x)×(x)]\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,
⇒tan⁡−1(2x2−x2)=tan⁡−1[2x1+3x2]\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,
⇒tan⁡[tan⁡−1(2x2−x2)]=tan⁡[tan⁡−1[2x1+3x2]]\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⁡(tan⁡−1x)=x\tan \left( {{\tan }^{-1}}x \right)=x
If we use the above formula in equation (2) we will get,
⇒2x2−x2=2x1+3x2\Rightarrow \dfrac{2x}{2-{{x}^{2}}}=\dfrac{2x}{1+3{{x}^{2}}}
By doing cross multiplication in the above equation we will get,
⇒2x(1+3x2)=2x(2−x2)\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,
⇒(1+3x2)=2x(2−x2)2x\Rightarrow \left( 1+3{{x}^{2}} \right)=\dfrac{2x\left( 2-{{x}^{2}} \right)}{2x}
By simplifying the above equation we will get,
⇒1+3x2=2−x2\Rightarrow 1+3{{x}^{2}}=2-{{x}^{2}}
If we shift the ‘−x2-{{x}^{2}}’ on the left hand side and 1 on the right hand side of the equation we will get,
⇒x2+3x2=2−1\Rightarrow {{x}^{2}}+3{{x}^{2}}=2-1
If we simplify the above equation we will get,
⇒4x2=1\Rightarrow 4{{x}^{2}}=1
Further simplification in the above equation will give,
⇒x2=14\Rightarrow {{x}^{2}}=\dfrac{1}{4}
Taking square roots on both sides of the above equation we will get,
⇒x=±12\Rightarrow x=\pm \dfrac{1}{2}

Therefore the solution of the equation tan⁡−1(x−1)+tan⁡−1x+tan⁡−1(x+1)=tan⁡−13x{{\tan }^{^{-1}}}(x-1)+{{\tan }^{^{-1}}}x+{{\tan }^{^{-1}}}(x+1)={{\tan }^{^{-1}}}3x is x=±12x=\pm \dfrac{1}{2}.
Therefore the correct answer is option (b).

Note: At the step tan⁡[tan⁡−1(2x2−x2)]=tan⁡[tan⁡−1[2x1+3x2]]\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{{\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.