Question

If $x^2+y^2+z^2=r^2$, then ${\tan }^{-1}\frac{xy}{zr}+{\tan }^{-1}\frac{yz}{xr}+{\tan }^{-1}\frac{zx}{yr}=$

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

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct option.

Sum of inverse trigonometric ratios:

$$\begin{gathered} {\tan }^{-1}\frac{xy}{zr}+{\tan }^{-1}\frac{yz}{xr}+{\tan }^{-1}\frac{zx}{yr} \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{xy}{zr}}+{\displaystyle \frac{yz}{xr}+\frac{zx}{yr}-\left(\frac{xy}{zr}\times \frac{yz}{xr}\times \frac{zx}{yr}\right)}}{1-\left({\displaystyle \frac{xy}{zr}}\times {\displaystyle \frac{yz}{xr}}\right)-\left({\displaystyle \frac{yz}{xr}}\times {\displaystyle \frac{zx}{yr}}\right)-\left({\displaystyle \frac{zx}{yr}}\times {\displaystyle \frac{xy}{zr}}\right)}\right]\left[\mathrm{by}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{C}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{+}\mathbf{C}\mathbf{-}\mathbf{ABC}}{\mathbf{1}\mathbf{-}\mathbf{A}\mathbf{B}\mathbf{-}\mathbf{BC}\mathbf{-}\mathbf{CA}}\mathbf{\right)}\right] \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{x^2{y}^2+y^2{z}^2+z^2{x}^2-x^2{y}^2{z}^2}{xyzr}}}{{\displaystyle \frac{r^2-\left(x^2+y^2+z^2\right)}{r^2}}}\right] \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{x^2{y}^2+y^2{z}^2+z^2{x}^2-x^2{y}^2{z}^2}{xyzr}}}{{\displaystyle 0}}\right]\left[\mathrm{Given}{x}^2+y^2+z^2=r^2\right] \\ ={\tan }^{-1}\infty \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, option B is correct.