First, let's simplify #1+cot^2(theta):#
Recall that #cot(theta)=cos(theta)/sin(theta)#, so it follows that #cot^2(theta)=cos^2(theta)/sin^2(theta)#
#1+cot^2(theta)=1+cos^2(theta)/sin^2(theta)=sin^2(theta)/sin^2(theta) + cos^2(theta)/sin^2(theta)=(sin^2(theta)+cos^2(theta))/sin^2(theta)#
Now, recall that #sin^2(theta)+cos^2(theta)=1#, so our numerator becomes #1#
#1/sin^2(theta)=csc^2(theta)#
Thus far, we have
#csc^2(theta)/(csc(theta)sec(theta))#
Simplify:
#csc^(cancel(2)1)(theta)/(cancel(csc(theta))sec(theta))#
#csc(theta)/sec(theta)#
Recall that #csc(theta)=1/sin(theta), sec(theta)=1/cos(theta)#
So,
#csc(theta)/sec(theta)=(1/sin(theta))/(1/cos(theta))=1/(sin(theta)/cos(theta))#
Because #(a/b)/c=a/(bc)#
#1/(sin(theta)/cos(theta))=cos(theta)/sin(theta)=cot(theta)#
because #1/(a/b)=b/a#
This proof falls in line with #c.#
