Question

$OA$ 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 $\Psi ={\Psi }_1+{\Psi }_2$

Answer 1

The bisector $OC$ is perpendicular to $AB$ as $△OAB$ is isosceles. $\angle AOB={\Psi }_2-{\Psi }_1$
$\therefore \angle COX=\frac{1}{2}({\Psi }_2-{\Psi }_1)+{\Psi }_1=\frac{{\Psi }_2+{\Psi }_1}{2}=\frac{\Psi }{2}$
Hence slope of bisector is
$\tan \frac{\Psi }{2}=\frac{\sin \frac{\Psi }{2}}{\cos \frac{\Psi }{2}}=\frac{1-\cos \Psi }{\sin \Psi }=co\sec \Psi -\cot \Psi$