How do you verify #tan^2(x) / sec(x) = sec(x) - cos(x) #?

1 Answer
John D.
Aug 22, 2015

First prove that #secx-cosx = (sinx)(tanx)#. Then, mess with the equation some to get the right answer.

Explanation:

(Lightly used source: https://answers.yahoo.com/question/index?qid=20080831084348AAj0BJV)

Start with the identity #sin^2x+cos^2x=1#

Subtract #cos^2x# from both sides:

#sin^2x = 1 - cos^2x#

Divide both sides by #cosx#.

#sin^2x/cosx = 1/cosx - cosx#

Now simplify. Remember sin/cos = tan, and 1/cos = sec

#sinx * sinx/cosx = secx - cosx#

#sinx*tanx = secx-cosx#

There. Now that we've proven that #(sinx)(tanx)=secx-cosx#, all we have to do is show that #tan^2x/secx = (sinx)(tanx)#.

#tan^2x/secx = tanx * tanx/secx = tanx*(tanx)(cosx) = tanx*sinx#

There we go! If this doesn't make sense, don't worry. Let me know in the 'comments' and I'll give you a list of the trigonometric identities I used to solve this.

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