How do you prove #(sec+tan)(sec-tan)=1#?
1 Answer
Jul 18, 2016
We will be using the following:
-
#(a+b)(a-b) = a^2-b^2# -
#tan^2(theta)+1 = sec^2(theta)#
With those:
#=sec^2(theta)-tan^2(theta)#
#=(tan^2(theta)+1)-tan^2(theta)#
#=1#
