Question

If $r^2=x^2+y^2+z^2$ and ${\tan }^{-1}\frac{yz}{xr}+{\tan }^{-1}\frac{xz}{yr}=\frac{\pi }{2}-{\tan }^{-1}\varphi$, then

• A. $\varphi =\frac{zr}{xy}$
• B. $\varphi =\frac{xy}{zr}$
• C. $\varphi =\frac{x+y}{zr}$
• D. $\varphi =\frac{yz}{xr}+\frac{xz}{yr}$

Answer 1

Answer: (B)

$\varphi =\frac{xy}{zr}$

We have, ${\tan }^{-1}\frac{yz}{xr}+{\tan }^{-1}\frac{xz}{yr}=\frac{\pi }{2}-{\tan }^{-1}\varphi$
Now, ${\tan }^{-1}\left(\frac{\frac{yz}{xr}+\frac{xz}{yr}}{1-\frac{yz}{xr}\cdot \frac{xz}{yr}}\right)=\frac{\pi }{2}-{\tan }^{-1}\varphi$
$[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)]$
$\Rightarrow {\tan }^{-1}\left[\frac{y^{2}zr+x^{2}zr}{xy{r}^2-xy{z}^2}\right]=\frac{\pi }{2}-{\tan }^{-1}\varphi$
$\Rightarrow {\tan }^{-1}\left[\frac{zr(x^2+y^2)}{xy(r^2-z^2)}\right]=\frac{\pi }{2}-{\tan }^{-1}\varphi$
$\Rightarrow {\tan }^{-1}\frac{zr(x^2+y^2)}{xy(x^2+y^2)}=\frac{\pi }{2}-{\tan }^{-1}\varphi$
$[\because x^2+y^2=r^2-z^2]$
$\Rightarrow {\tan }^{-1}\frac{zr}{xy}={\cot }^{-1}\varphi$ $[\because {\tan }^{-1}\theta +{\cot }^{-1}\theta =\frac{\pi }{2}]$
$\Rightarrow {\tan }^{-1}\frac{zr}{xy}={\tan }^{-1}\frac{1}{\varphi }$ $[\because {\tan }^{-1}\frac{1}{\theta }={\cot }^{-1}\theta ]$
$\Rightarrow \frac{1}{\varphi }=\frac{zr}{xy}\Rightarrow \varphi =\frac{xy}{zr}$.