Question

If $x^2+y^2+z^2=r^2,$ then ${\tan }^{-1}\left(\frac{xy}{zr}\right)+{\tan }^{-1}\left(\frac{yz}{xr}\right)+{\tan }^{-1}\left(\frac{zx}{yr}\right)$ is equal to

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. 0
• D. none of these

Answer 1

Answer: (B)

${\tan }^{-1}\left(\frac{xy}{zr}\right)+{\tan }^{-1}\left(\frac{yz}{xr}\right)+{\tan }^{-1}\left(\frac{zx}{yr}\right)$
$={\tan }^{-1}\left[\frac{\frac{xy}{zr}+\frac{yz}{xr}+\frac{zx}{yr}-\frac{xyz}{r^3}}{1-\left(\frac{x^2+y^2+z^2}{r^2}\right)}\right]$
$={\tan }^{-1}\infty =\frac{\pi }{2}\left(\because x^2+y^2+z^2=r^2\right)$