Expert: Sherry Wallin - 6/4/2009

Question
I Need help establishing an identity:
Sec-Cos/Sec+Cos  =    Sin^2/1+cos^2

Answer
Hi Halen~
    First we need to correct the notation. The trig functions are functions and that means they need to have an argument.Normally you see theta or x as in [secx-cosx]/[secx+cosx] = sin^2x/(1+cos^2x). Please notice also the use of parentheses or brackets, this is necessary to communicate exactly what the author intended when you type in the problem.

There is a thing called the conjugate of a sum or difference. If you have a sum then it is the difference,if you have a difference then it is a sum.(a+b) is a sum so (a-b) is a difference. (x-2) is a difference so (x+2) is a sum. We use conjugates to simplify things and in particular to clean up things in the denominators. When you multiply a sum and difference that is a real number you always gets the first guy squared minus the second guy scared ignoring signs.
So (a+b)(a-b) = a^2 - b^2 Let's multiply the top and bottom of the left hand side by secx-cosx getting:
(secx-cosx)(secx-cosx)/(secx+cosx)(secx-cosx)
= (sec^2x-2cosxsecx+cos^2x)/(sec^2x-cos^2x)
because cosx*secx = cosx*(1/cosx) = 1
= (sec^2x-2+cos^2x)/(sec^2x-cos^2x)
= [(1/cos^2x)-2+cos^2x]/[(1/cos^2x)-cos^2x] because sec^2x = 1/cos^2x
now multiply top and bottom by cos^2x getting:
= [1-2cos^2x+cos^4x]/[1-cos^4x] which just happens to factor as:
= ((1-cos^2x)(1-cos^2x)]/(1+cos^2x)(1-cos^2x) cancel common factors:
= (1-cos^2x)/(1+cos^2x)
= sin^2x/(1+cos^2x) because sin^2x = 1 - cos^2x

Math Prof