Question

If $x^2+y^2+z^2=r^2$ and $\tan \alpha =\frac{xy}{zr},\tan \beta =\frac{yz}{xr},\tan \gamma =\frac{ZX}{yr}$ then $\alpha +\beta +\gamma =$

• A. $\pi /4$
• B. $\pi$
• C. $\pi /2$
• D. $\pi /3$

Answer 1

The correct option is C $\pi /2$

Given $x^2+y^2+z^2=r^2$

$tan\alpha =\frac{xy}{zr},tan\beta =\frac{yz}{xr},tan\gamma =\frac{zx}{yr}$
$\therefore tan\alpha tan\beta tan\gamma =\frac{xyz}{r^3}$
$\therefore tan(\alpha +\beta +\gamma )=\frac{tan\alpha +tan\beta +tan\gamma -tan\alpha tan\beta tan\gamma }{1-tan\alpha tan\beta -tan\beta tan\gamma -tan\alpha tan\gamma }$
$=\frac{\frac{xy}{zr}+\frac{yz}{xr}+\frac{zx}{yr}-\frac{xyz}{r^3}}{1-\frac{y^2}{r^2}-\frac{z^2}{r^2}-\frac{x^2}{r^2}}$
$=\frac{(\frac{xy}{zr}+\frac{yz}{xr}+\frac{xz}{yr}-\frac{xyz}{r^3})}{1-\left(1\right)}$
$\therefore tan(\alpha +\beta +\gamma )=\infty$
$\Rightarrow \alpha +\beta +\gamma =\frac{\pi }{2}$