Question

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 $x^4-x^2\left(\sum ab\right)+abcd=$

• A. $-1$
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

${\tan }^{-1}\frac{a}{x}+{\tan }^{-1}\frac{b}{x}+{\tan }^{-1}\frac{c}{x}+{\tan }^{-1}\frac{d}{x}=\frac{\pi }{2}$
$\Rightarrow {\tan }^{-1}\frac{a}{x}+{\tan }^{-1}\frac{b}{x}=\frac{\pi }{2}-\left\{{\tan }^{-1}\frac{c}{x}+{\tan }^{-1}\frac{d}{x}\right\}$
$\Rightarrow {\tan }^{-1}\left(\frac{\frac{a}{x}+\frac{b}{x}}{1-\frac{ab}{x^2}}\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\frac{c}{x}+\frac{d}{x}}{1-\frac{cd}{x^2}}\right)$
$\Rightarrow {\tan }^{-1}\frac{(a+b)x}{x^2-ab}=\frac{\pi }{2}-{\tan }^{-1}\frac{(c+d)x}{x^2-cd}$
$\Rightarrow {\tan }^{-1}\frac{(a+b)x}{x^2-ab}={\cot }^{-1}\frac{(c+d)x}{x^2-cd}$
$\Rightarrow {\tan }^{-1}\frac{(a+b)x}{x^2-ab}={\tan }^{-1}\frac{x^2-cd}{(c+d)x}$
$\Rightarrow \frac{(a+b)x}{x^2-ab}=\frac{x^2-cd}{(c+d)x}$
$\Rightarrow x^4-x^2\left(\sum ab\right)+abcd=0$