Show that (1 - cos(2x)) / (1 + cos(2x)) = sec^2(x) - 1

First, take the side of the equation that looks most complicated because it often needs simplifying. This is the LHS in this case. The LHS has cos(2x) twice - therefore the double angle formula probably applies:cos(2x) = cos^2(x) - sin^2(x) = 2cos^2(x) - 1 = 1 - 2sin^2(x)In this case the cos(2x) = 2cos^2(x) - 1 looks most useful as we have sec^2(x) on the RHS of our equation and cos(x) = 1/sec(x).Sub the double angle formula into LHS:(1 - (2cos^2(x) - 1)) / (1 + (2cos^2(x) - 1)= (2 - 2cos^2(x)) / (2cos^2(x))= 2/2cos^2(x) - 2cos^2(x)/2cos^2(x)= sec^2(x) - 1= RHS.Therefore (1 - cos(2x)) / (1 + cos(2x)) = sec^2(x) - 1