Question

Solve for x - tan^-1 (x+1) + tan^-1 (x-1) = tan^-1 8/31

Answer 1

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Answer 2

Answer:

The equation can be solved as follows.

$\tan ^{-1}\left\{\frac{x+1+x-1}{1-(x+1)(x-1)}\right\}=\tan ^{-1}\left(\frac{8}{31}\right)$ (-1 and 1 gets cancelled)

$\tan ^{-1}\left\{\frac{2 x}{1-\left(x^{2}-1\right)}\right\}=\tan -1\left(\frac{8}{31}\right)$

$\tan ^{-1}\left[\frac{2 x}{2-x^{2}}\right]=\tan ^{-1}\left(\frac{8}{31}\right)$

$\tan ^{-1}$ get cancelled both sides,

Therefore,  

$2 x /\left(2-x^{2}\right)=8 / 31$

On cross multiplying,

$\begin{array}{l}{2 x \times 31=8 \times\left(2-x^{2}\right)} \\ {62 x=16-8 x^{2}} \\ {8 x^{2}+62 x-16=0}\end{array}$

4 x^{2}+31 x-8=0 Taking quadratic equation

(x+8) (4x-1) =0   we get the roots as

x=-8   and x=1/4

Since x=-8 is negative

Hence the only solution is x=1/4.