We know the distance formula:
The distance between color(Brown)(P(x_1,y_1) and Q(x_2,y_2) is:
color(Brown)(PQ=sqrt((x_1-x_2)^2+(y_1-y_2)^2)...to(D)
For simplicity we take ,(theta to x)
A(csc^2x,0) ,B(0,sec^2x) and C(1,1)
Usingcolor(brown)( (D) we get
(AB)^2=(csc^2x-0)^2+(0-sec^2x)^2
color(white)((AB)^2)=(1+cot^2x)^2+(1+tan^2x)^2
color(white)((AB)^2)=(1+1/tan^2x)^2+(1+tan^2x)^2
color(white)((AB)^2)=(tan^2x+1)^2/tan^4x+(1+tan^2x)^2/1
color(white)((AB)^2)=(1+tan^2x)^2[1/tan^4x+1]
color(white)((AB)^2)=(sec^2x)^2[(1+tan^4x)/tan^4x]
:.(AB)^2=(sec^2x/tan^2x)^2[1+tan^4x]
=>color(red)(AB=sec^2x/tan^2xsqrt(1+tan^4x)...to(I)
Again usingcolor(brown)( (D) we get
(AC)^2=(csc^2x-1)^2+(0-1)^2
=>(AC)^2=(cot^2x)^2+1
=>(AC)^2=1/tan^4x+1=(1+tan^4x)/tan^4x
=>(AC)^2=(1+tan^4x)/(tan^2x)^2
=>color(blue)(AC=sqrt(1+tan^4x)/tan^2x...to(II)
Again usingcolor(brown)( (D) we get
(CB)^2=(0-1)^2+(sec^2x-1)^2=1+(tan^2x)^2
=>(CB)^2=1+tan^4x
=>color(blue)(CB=sqrt(1+tan^4x)...to(III)
Adding (II) and(III) we get
AC+CB=sqrt(1+tan^4x)/tan^2x+sqrt(1+tan^4x)/1
=>AC+CB=sqrt(1+tan^4x)[1/tan^2x+1]
=>AC+CB=sqrt(1+tan^4x)[(1+tan^2x)/tan^2x]
=>AC+CB=sqrt(1+tan^4x)[sec^2x/tan^2x] or
color(red)(AC+CB=sec^2x/tan^2xsqrt(1+tan^4x)...to(IV)
From (I) and (IV) we can say that
color(violet)(AC+CB=AB=>A ,B ,C "are "color(violet)"collinear points. "
Note:
(i)csc^2theta-cot^2theta=1=>csc^2theta-1=cot^2theta
(ii)sec^2theta-tan^2theta=1=>sec^2theta-1=tan^2theta