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

Hint: 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{align}
  & \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{align} $
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{align}
  & 62x=8[1-{{x}^{2}}+1] \\
 & \Rightarrow 62x=8[2-{{x}^{2}}] \\
 & \Rightarrow 62x=16-8{{x}^{2}} \\
\end{align} $
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{align}
  & 4{{x}^{2}}+32x-x-8=0 \\
 & \Rightarrow 4x(x+8)-1(x+8)=0 \\
 & \Rightarrow (4x-1)(x+8)=0 \\
\end{align} $
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.