$OA$ and $OB$ are equal sides of an isosceles triangle lying in the first quadrant, where $O$ is the origin. $OA$ and $OB$ make angles ${ \Psi }_{ 1 }$ and${ \Psi }_{ 2 }$ respectively with the +ive axis. Show that the slope of the bisector of the acute angle $AOB$ is $co\sec { \Psi } -\cot { \Psi }$ where $\Psi ={ \Psi }_{ 1 }+{ \Psi }_{ 2 }\quad$

#### Solution

The bisector $OC$ is perpendicular to $AB$ as $\triangle {OAB}$ is isosceles. $\angle AOB={ \Psi }_{ 2 }-{ \Psi }_{ 1 }$
$\therefore \angle COX=\cfrac { 1 }{ 2 } \left( { \Psi }_{ 2 }-{ \Psi }_{ 1 } \right) +{ \Psi }_{ 1 }=\cfrac { { \Psi }_{ 2 }+{ \Psi }_{ 1 } }{ 2 } =\cfrac { \Psi }{ 2 } \quad$
Hence slope of bisector is
$\tan { \cfrac { \Psi }{ 2 } } =\cfrac { \sin { \cfrac { \Psi }{ 2 } } }{ \cos { \cfrac { \Psi }{ 2 } } } =\cfrac { 1-\cos { \Psi } }{ \sin { \Psi } } =co\sec { \Psi } -\cot { \Psi }$