$$OA$$ 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 
$$\tan { \cfrac { \Psi  }{ 2 }  } =\cfrac { \sin { \cfrac { \Psi  }{ 2 }  }  }{ \cos { \cfrac { \Psi  }{ 2 }  }  } =\cfrac { 1-\cos { \Psi  }  }{ \sin { \Psi  }  } =co\sec { \Psi  } -\cot { \Psi  } $$