Question

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

• A. ${\sum }_{i=1}^4\left(x_i\right)=a+b+c+d$
• B. ${\sum }_{i=1}^4\left(\frac{1}{x_i}\right)=0$
• C. ${\prod }_{i=1}^4\left(\frac{1}{x_i}\right)=0$
• 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), (D)

The correct options are
B

${\sum }_{i=1}^4\left(\frac{1}{x_i}\right)=0$

D

$(x_1+x_2+x_3)(x_2+x_3+x_4)(x_3+x_4+x_1)(x_4+x_1+x_2)=abcd$

Let $ta{n}^{-1}\left(\frac{a}{x}\right)=\alpha \Rightarrow tan\alpha =\frac{a}{x}$ and so on.

$\Rightarrow tan(\alpha +\beta +\gamma +\delta )=tan\left(\frac{\pi }{2}\right)$

$\Rightarrow \frac{S_1-S_3}{1-S_2+S_4}=\infty \Rightarrow 1-S_2+S_4=0\Rightarrow S_4-S_2+1=0$

Now, $S_4=\left(tan\alpha \right)\left(tan\beta \right)\left(tan\gamma \right)\left(tan\delta \right)=\frac{abcd}{x^4}$

$S_2=\sum \left(tan\alpha tan\beta \right)=\sum \left(\frac{ab}{x^2}\right)$

$\therefore \frac{abcd}{x^4}-\frac{\sum ab}{x^2}+1=0$

$\Rightarrow x^4-\sum \left(ab\right)x^2+abcd=0$

Its roots are $x_1,x_2,x_3,x_4.$

$\therefore x_1+x_2+x_3+x_4=0$ ....(1)

$\sum \left(x_1{x}_2{x}_3\right)=x_1{x}_2{x}_3{x}_4[\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}]=0$

Also, $x_1{x}_2{x}_3{x}_4=abcd.$

$(x_1+x_2+x_3)(x_2+x_3+x_4)(x_3+x_4+x_1)(x_4+x_1+x_2)=(\sum x_i-x_1)(\sum x_i-x_2)(\sum x_i-x_3)(\sum x_i-x_4)=x_1{x}_2{x}_3{x}_4=abcd$