Question

Let $x_1,x_2,x_3,x_4$ be four non zero numbers satisfying the equation
${\tan }^{-1}\frac{a}{x}+{\tan }^{-1}\frac{b}{x}+{\tan }^{-1}\frac{c}{x}+{\tan }^{-1}\frac{d}{x}=\frac{\pi }{2}$
Then, which of the following relation(s) hold good?

• A. $\sum _{i=1}^4{x}_i=a+b+c+d$
• B. $\sum _{i=1}^4\frac{1}{x_i}=0$
• C. $\prod _{i=1}^4{x}_i=abcd$
• D. $(x_1+x_2+x_3)(x_2+x_3+x_4)(x_3+x_4+x_1)(x_4+x_1+x_2)=abcd$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\sum _{i=1}^4\frac{1}{x_i}=0$
C $\prod _{i=1}^4{x}_i=abcd$
D $(x_1+x_2+x_3)(x_2+x_3+x_4)(x_3+x_4+x_1)(x_4+x_1+x_2)=abcd$

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

Let ${\tan }^{-1}\frac{a}{x}=\alpha ,{\tan }^{-1}\frac{b}{x}=\beta ,{\tan }^{-1}\frac{c}{x}=\gamma ,{\tan }^{-1}\frac{d}{x}=\delta$
Now,
$\alpha +\beta +\gamma +\delta =\frac{\pi }{2}\Rightarrow \tan (\alpha +\beta +\gamma +\delta )=\tan \frac{\pi }{2}\Rightarrow \frac{S_1-S_3}{1-S_2+S_4}=\frac{1}{0}\Rightarrow 1-S_2+S_4=0$

Where,
$S_4=\tan \alpha \cdot \tan \beta \cdot \tan \gamma \cdot \tan \delta \Rightarrow S_4=\frac{abcd}{x^4}{S}_2=\sum \tan \alpha \cdot \tan \beta \Rightarrow S_2=\frac{\sum ab}{x^2}$

So,
$1-S_2+S_4=0\Rightarrow 1-\frac{\sum ab}{x^2}+\frac{abcd}{x^4}=0\Rightarrow x^4-\left(\sum ab\right)x^2+abcd=0$
Whose roots are $x_1,x_2,x_3,x_4$, so
$x_1+x_2+x_3+x_4=0...\left(1\right)$

$x_1{x}_2{x}_3{x}_4=abcd$

$\sum x_1{x}_2{x}_3=0$

$\sum _{i=1}^4\frac{1}{x_i}=\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\Rightarrow \sum _{i=1}^4\frac{1}{x_i}=\frac{\sum x_1{x}_2{x}_3}{x_1{x}_2{x}_3{x}_4}=0$

$(x_1+x_2+x_3)(x_2+x_3+x_4)(x_3+x_4+x_1)(x_4+x_1+x_2)=(-x_4)(-x_1)(-x_2)(-x_3)\left[\text{From}\right(1\left)\right]=x_1{x}_2{x}_3{x}_4=abcd$