Question

tan^-1(yz/xr)+tan^-1(zx/yr)+tan^-1(xy/zr)=pie/4 Prove that x^2+y^2+z^2=r^2

Answer 1

tan-1(x) + tan-1(y)= tan-1 (x+y/1-xy)
take 2 at a time and solve.
tan-1(1)=pie/4

Answer 2

Answer:

The value of $x^2+y^2+z^2=r^2$. Hence proved the given expression.

To prove:

The value of $x^2+y^2+z^2=r^2$

Solution:

Given that

$\tan ^ { - 1 } \left( \frac { \mathrm { yz } } { \mathrm { xr } } \right) + \tan ^ { - 1 } \left( \frac { \mathrm { zx } } { \mathrm { yr } } \right) + \tan ^ { - 1 } \left( \frac { \mathrm { xy } } { \mathrm { zr } } \right) = \frac { \pi } { 4 }$

$\tan ^ { - 1 } \mathrm { m } + \tan ^ { - 1 } \mathrm { n } = \tan ^ { - 1 } \left( \frac { \mathrm { m } + \mathrm { n } } { 1 - \mathrm { mn } } \right)$

$\tan ^ { - 1 } \left( \frac { y z } { x r } \right) + \tan ^ { - 1 } \left( \frac { z x } { y r } \right) = \tan ^ { - 1 } \left( \frac { \frac { y z } { x r } + \frac { z x } { y r } } { 1 - \frac { y z } { x r } \times \frac { z x } { y r } } \right)$

$\begin{array} { l } { = \tan ^ { - 1 } \left( \frac { \mathrm { y } ^ { 2 } \mathrm { zr } + \mathrm { zx } ^ { 2 } \mathrm { r } } { \mathrm { xyr } ^ { 2 } - \mathrm { yz } ^ { 2 } \mathrm { x } } \right) } \\\\ { = \tan ^ { - 1 } \left\{ \frac { \mathrm { zr } \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } \right) } { \mathrm { xy } \left( \mathrm { r } ^ { 2 } - \mathrm { z } ^ { 2 } \right) } \right\} } \end{array}$

$\tan ^ { - 1 }\frac {zr(x^2+y^2)}{xy(r^2-z^2)}+\tan ^ { - 1 } \frac {(xy)}{(zr)}=\frac {\pi}{4}$

$\tan^{-1} \frac {zr(x^2+y^2)}{xy(r^2-z^2)}= \frac {\pi}{4}-\tan^{-1}\frac{xy}{zr}$

Taking tan on both sides, We will get,

$\frac {zr(x^2+y^2)}{xy(r^2-z^2)} =\tan {(\frac {\pi}{4}-\tan^{-1} \frac {xy}{zr})}$

$\tan \left( \frac { \pi } { 2 } - A \right) = \cot A$

$\frac {zr(x^2+y^2)}{xy(r^2-z^2)} =\cot {(\tan^{-1} \frac {xy}{zr})}\\\\\frac {zr(x^2+y^2)}{xy(r^2-z^2)} =\tan {(\tan^{-1} {\frac {zr}{xy}})}$

We know that the value of

$cot {(\tan^{-1} (x))}=\cot {(\cot^{-1} \frac {1}{x})}\\\\\frac {zr(x^2+y^2)}{xy(r^2-z^2)} = \frac {zr}{xy}$

Cancel the zr and xy on both sides, we will get,

$\frac {((x^2+y^2))}{((r^2-z^2))}=1\\\\x^2+y^2=r^2-z^2\\\\x^2+y^2+z^2=r^2$

Hence proved