How do you verify #(sec + tan)(sec - tan) = 1#?

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Answer 1 · 2016-07-15T06:16:30

#cos^2 alpha/cos^2 alpha=1#

Since

#(a+b)(a-b)=a^2-b^2#

you will have:

#(sec alpha+tan alpha)(sec alpha-tan alpha)=sec^2 alpha-tan^2 alpha#

then, since

#sec alpha=1/cos alpha#

and

#tan alpha=sin alpha/cos alpha#

you will have:

#1/cos^2 alpha-sin^2alpha/cos^2 alpha=#

#(1-sin^2 alpha)/cos^2 alpha=#

#cos^2 alpha/cos^2 alpha=1#