Solve for `x :` `tan^(-1)(x+2)+tan^(-1)(x-2)=tan^(-1)(8/(79))` , `x >0`

Answer

`x=1/4`

Explanation

\begin{aligned}
&\text { We know }\\
&\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\\
&\therefore \tan ^{-1}(x+2)+\tan ^{-1}(x-2)=\tan ^{-1}\left(\frac{8}{79}\right)\\
&\Rightarrow \tan ^{-1}\left(\frac{x+2+x-2}{1-(x+2) \times(x-2)}\right)=\tan ^{-1} \frac{8}{79}\\
&\Rightarrow \frac{2 x}{1-x^{2}+4}=\frac{8}{79}\\
&\Rightarrow \frac{x}{5-x^{2}}=\frac{4}{79}\\
&\Rightarrow 79 x=20-4 x^{2}\\
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