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)#
Using#color(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 using#color(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 using#color(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#
