Problem 0003
If r2 = x2 + y2 + z2, then prove that
[tex]{\tan ^{ - 1}}\left( {{{yz} \over {rx}}} \right) + {\tan ^{ - 1}}\left( {{{zx} \over {ry}}} \right) + {\tan ^{ - 1}}\left( {{{xy} \over {rz}}} \right) = {\pi \over 2}[/tex]
Answer:
[tex]LHS = {\tan ^{ - 1}}\left( {{{xyz} \over {r{x^2}}}} \right) + {\tan ^{ - 1}}\left( {{{xyz} \over {r{y^2}}}} \right) + {\tan ^{ - 1}}\left( {{{xyz} \over {r{z^2}}}} \right)[/tex]
Let
[tex]{{xyz} \over r} = p[/tex]
There fore
[tex] = {\tan ^{ - 1}}\left( {{p \over {{x^2}}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{y^2}}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex]{\tan ^{ - 1}}\left( {{{{p \over {{x^2}}} + {p \over {{y^2}}}} \over {1 - {p^2}/({x^2}{y^2})}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\tan ^{ - 1}}\left( {{{({x^2} + {y^2})k} \over {{x^2}{y^2} - {k^2}}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\tan ^{ - 1}}\left( {{{({r^2} - {z^2})k} \over {{x^2}{y^2} - {k^2}}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\tan ^{ - 1}}\left( {{{({r^2} - {z^2})k{z^2}} \over {{x^2}{y^2}{z^2} - {k^2}{z^2}}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\tan ^{ - 1}}\left( {{{({r^2} - {z^2})k{z^2}} \over {{k^2}{r^2} - {k^2}{z^2}}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\tan ^{ - 1}}\left( {{{({r^2} - {z^2}){z^2}} \over {k({r^2} - {z^2})}}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\tan ^{ - 1}}\left( {{{{z^2}} \over k}} \right) + {\tan ^{ - 1}}\left( {{p \over {{z^2}}}} \right)[/tex]
[tex] = {\pi \over 2}[/tex]
