If $tan^{-1}\frac{a}{x} + tan^{-1}\frac{b}{x}+tan^{-1}\frac{c}{x} +tan^{-1}\frac{d}{x}=\frac{\pi}{2}, $ then $ x^4 - x^2 (∑ab) + abcd = $

0

We have,

$tan^{-1}\frac{a}{x} + tan^{-1}\frac{b}{x}+tan^{-1}\frac{c}{x} +tan^{-1}\frac{d}{x}=\frac{\pi}{2}$

$⇒ tan^{-1}\frac{a}{x} + tan^{-1}\frac{b}{x} =\frac{\pi}{2}- \begin{Bmatrix}tan^{-1}\frac{c}{x} +tan^{-1}\frac{d}{x}\end{Bmatrix}$

$⇒ tan^{-1}\left(\frac{\frac{a}{x}+\frac{b}{x}}{1-\frac{ab}{x^2}}\right) = \frac{\pi}{2}- tan^{-1}\left(\frac{\frac{c}{x}+\frac{d}{x}}{1-\frac{cd}{x^2}}\right)$

$⇒ tan^{-1} \frac{(a+b)x}{x^2-ab}=\frac{\pi}{2}-tan^{-1} \frac{(c+d)x}{x^2-cd}$

$⇒ tan^{-1} \frac{(a+b)x}{x^2-ab}=cot^{-1} \frac{(c+d)x}{x^2-cd}$

$⇒ tan^{-1} \frac{(a+b)x}{x^2-ab}=tan^{-1} \frac{x^2-cd}{(c+d)x}$

$⇒\frac{(a+b)x}{x^2-ab} = \frac{x^2-cd}{(c+d)x} ⇒ x^4 - x^2 (∑ab) + abcd = 0 $