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)

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