Given:
#sec(theta)/(csc(theta)-cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#
Multiply the first fraction by 1 in the form #(csc(theta)+cot(theta))/(csc(theta)+cot(theta))#
#sec(theta)/(csc(theta)-cot(theta))(csc(theta)+cot(theta))/(csc(theta)+cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#
This makes the denominator become the difference of two squares:
#(sec(theta)(csc(theta)+cot(theta)))/(csc^2(theta)-cot^2(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#
This particular difference of two squares is the left side of the identity #csc^2(theta)-cot^2(theta) = 1#, therefore, the denominator becomes 1 and disappears:
#sec(theta)(csc(theta)+cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#
Use the identities #sec(theta)csc(theta) =1# and #sec(theta)cot(theta) = csc(theta)#:
#1+csc(theta)-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#
Multiply the next fraction by 1 in the form #(csc(theta)-cot(theta))/(csc(theta)-cot(theta))#
#1+csc(theta)-sec(theta)/(csc(theta)+cot(theta))(csc(theta)-cot(theta))/(csc(theta)-cot(theta)) = 2csc(theta)#
The produces the same difference of two squares, therefore, we shall merely delete the numerators:
#1+csc(theta)-sec(theta)(csc(theta)-cot(theta)) = 2csc(theta)#
Use the identities #sec(theta)csc(theta) =1# and #sec(theta)cot(theta) = csc(theta)#:
#1+csc(theta)-1+csc(theta) = 2csc(theta)#
Combine like terms:
#2csc(theta) = 2csc(theta)# Q.E.D.