Question

$\sum ta{n}^{-1}\left(\frac{yz}{x\sqrt{x^2+y^2+z^2}}\right)=?$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{2}$

Let
$A=\frac{yz}{x\sqrt{x^2+y^2+z^2}},B=\frac{zx}{y\sqrt{x^2+y^2+z^2}},C=\frac{xy}{z\sqrt{x^2+y^2+z^2}}$
then ,
$\Rightarrow ta{n}^{-1}A+ta{n}^{-1}B+ta{n}^{-1}C$

$ta{n}^{-1}\left(\frac{A+B}{1-AB}\right)+ta{n}^{-1}C$

$\Rightarrow ta{n}^{-1}\left(\frac{\frac{A+B}{1-AB}+C}{1-\frac{(A+B)C}{1-AB}}\right)$
$\Rightarrow ta{n}^{-1}\left(\frac{A+B+C-ABC}{1-(AB+BC+CA)}\right)$
$AB=\frac{z^2}{x^2+y^2+z^2}$
$BC=\frac{x^2}{x^2+y^2+z^2}$
$CA=\frac{y^2}{x^2+y^2+z^2}$
$\Rightarrow AB+BC+CA=1$
Denominator $=1-1=0$
$ta{n}^{-1}\left(\infty \right)=\frac{\pi }{2}$