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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 Ψ1 andΨ2 respectively with the +ive axis. Show that the slope of the bisector of the acute angle AOB is cosecΨcotΨ where Ψ=Ψ1+Ψ2

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Solution

The bisector OC is perpendicular to AB as OAB is isosceles. AOB=Ψ2Ψ1
COX=12(Ψ2Ψ1)+Ψ1=Ψ2+Ψ12=Ψ2
Hence slope of bisector is
tanΨ2=sinΨ2cosΨ2=1cosΨsinΨ=cosecΨcotΨ
1102417_1007340_ans_633f960461f543e4b1d259e4e45cddea.png

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