Question

The given equation is ${{\tan }^{-1}}(x+1)+{{\tan }^{-1}}(x-1)={{\tan }^{-1}}\left( \dfrac{8}{31} \right)$ find the value of x.

Answer 1

Now we know that ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$ . Hence we will simplify the left hand side with this formula. Then we will use the formula that ${{\tan }^{-1}}A={{\tan }^{-1}}B\Rightarrow A=B$
Hence we will find an equation in x. After simplifying the equation we will get a quadratic in x. We will solve the quadratic to find the values of x.

Complete step-by-step answer:
Now consider the given equation ${{\tan }^{-1}}(x+1)+{{\tan }^{-1}}(x-1)={{\tan }^{-1}}\left( \dfrac{8}{31} \right)$
Now we know that ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$ hence using this we get
${{\tan }^{-1}}\left( \dfrac{x+1+x-1}{1-(x+1)(x-1)} \right)={{\tan }^{-1}}\left( \dfrac{8}{31} \right)$
Now we know that if ${{\tan }^{-1}}A={{\tan }^{-1}}B\Rightarrow A=B$
Hence we will get the equation
$\begin{aligned} & \left( \dfrac{x+1+x-1}{1-(x+1)(x-1)} \right)=\left( \dfrac{8}{31} \right) \\\ & \Rightarrow \dfrac{2x}{1-(x+1)(x-1)}=\dfrac{8}{31} \\\ \end{aligned}$
Now we know that $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$ using this formula we get.
$\left( \dfrac{2x}{1-({{x}^{2}}-1)} \right)=\left( \dfrac{8}{31} \right)$
Cross multiplying the above equation we get
$31(2x)=8[1-({{x}^{2}}-1)]$
Now let us open the brackets
$\begin{aligned} & 62x=8[1-{{x}^{2}}+1] \\\ & \Rightarrow 62x=8[2-{{x}^{2}}] \\\ & \Rightarrow 62x=16-8{{x}^{2}} \\\ \end{aligned}$
Now taking all the terms of RHS to LHS we get
$8{{x}^{2}}+62x-16=0$
Dividing the equation throughout by 2 we get
$4{{x}^{2}}+31x-8=0$
Now we can write 31x as 32x – x . Hence we get
$\begin{aligned} & 4{{x}^{2}}+32x-x-8=0 \\\ & \Rightarrow 4x(x+8)-1(x+8)=0 \\\ & \Rightarrow (4x-1)(x+8)=0 \\\ \end{aligned}$
Now we know that $a.b=0\Rightarrow a=0\text{ or }b=0$
Hence we get $4x-1=0$ or $x+8=0$
Now if we consider 4x – 1 = 0
We get 4x = 1 which means $x=\dfrac{1}{4}$
Similarly if we consider x + 8 = 0
Then we get x = - 8
Now if we take x = - 8
${{\tan }^{-1}}(-8+1)+{{\tan }^{-1}}(-8-1)={{\tan }^{-1}}(-7)+{{\tan }^{-1}}(-9)$
Hence by applying the formula ${{\tan }^{-1}}A+{{\tan }^{-1}}B$ is given by ${{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$ we get
${{\tan }^{-1}}\left( \dfrac{-7-9}{1-(-7)(-9)} \right)={{\tan }^{-1}}\left( \dfrac{-16}{1-63} \right)={{\tan }^{-1}}\left( \dfrac{-16}{-62} \right)={{\tan }^{-1}}\left( \dfrac{8}{31} \right)$
Hence x = -8 does not satisfy the equation.
Now if we take $x=\dfrac{1}{4}$
${{\tan }^{-1}}(\dfrac{1}{4}+1)+{{\tan }^{-1}}(\dfrac{1}{4}-1)={{\tan }^{-1}}\left( \dfrac{5}{4} \right)+{{\tan }^{-1}}\left( -\dfrac{3}{4} \right)$
Hence by applying the formula ${{\tan }^{-1}}A+{{\tan }^{-1}}B$ is given by ${{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$ we get
${{\tan }^{-1}}\left( \dfrac{\dfrac{5}{4}+\left( \dfrac{-3}{4} \right)}{1-\left( \dfrac{5}{4} \right)\left( \dfrac{-3}{4} \right)} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{4}}{1+\dfrac{15}{16}} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{16}{2}}{16+15} \right)={{\tan }^{-1}}\left( \dfrac{8}{31} \right)$

Hence the values of x = $\dfrac{1}{4}$ and x = -8 both satisfy the equation.

Note: The formula for ${{\tan }^{-1}}A+{{\tan }^{-1}}B$ is given by ${{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$ and the formula of ${{\tan }^{-1}}A-{{\tan }^{-1}}B$ is given by ${{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)$ note that you take signs properly.