Solve for x:
\tan ^{-1}\left ( x+1 \right )+\tan ^{-1}\left ( x-1 \right )= \tan ^{-1}\left ( \frac{8}{31} \right )

 

 

 

 
 
 
 
 

Answers (1)

given: \tan^{-1}\left ( x+1 \right )+\tan^{-1}\left ( x-1 \right )= \tan^{-1}\frac{8}{13}
we know that
\tan^{-1}x+\tan^{-1}y= \tan^{-1}\frac{x+y}{1-xy}, xy<1
\therefore \tan^{-1}\left ( x+1 \right )+\tan^{-1}\left ( x-1 \right )= \tan^{-1}\left [ \frac{x+1+x-1}{1-\left [ \left ( x+1 \right )\left ( x-1 \right ) \right ]} \right ]
                                                                    \Rightarrow \tan^{-1}\left [ \frac{2x}{1-\left ( x^{2}-1 \right )} \right ]
                                                                    \Rightarrow \tan^{-1}\left [ \frac{2x}{2-x^{2}} \right ]
Now,\Rightarrow \tan^{-1}\frac{2x}{2-x^{2}}= \tan^{-1}\frac{8}{31}
       \Rightarrow \frac{2x}{2-x^{2}}= \frac{8}{31}\Rightarrow 62x= 16-8x^{2}
      \Rightarrow 31x= 8-4x^{2}\Rightarrow 4x^{2}+31x-8= 0
     \Rightarrow 4x^{2}+32x-x-8= 0\: \Rightarrow 4x\left ( x+8 \right )-1\left ( x+8 \right )= 0
    \Rightarrow \left ( x+8 \right )\left ( 4x-1 \right )= 0 \: \Rightarrow x= -8\: and \: \: x= \frac{1}{4}

 xy<1 \Rightarrow (x+1)(x-1)<1 \\ x^2-1<1 \\ x^2<2

So only answer is x=\frac{1}{4}
       

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