Oh now I see your point. I am sorry I should have stated this before. How do I verify that tan^2(theta)/Sec(theta)-1 is somehow the same as 1+cos(theta)/Cos(Theta)?

Answer 1

I went step by step, transforming the left side to the point where it equaled the right side.

I started with the left side as written, made substitutions which did not change its value, and ended up with the right side.

Looks like you need to review some of the proofs that you've already seen. You make substitutions until one side is the same as the other.

We could have worked with both sides, such as

(1+cos)/cos = tan^2/(sec-1)
clear fractions by cross-multiplying
(1+cos)(sec-1) = tan^2 * cos
expand both sides
sec + 1 - 1 - cos = sin^2/cos
sec-cos = sin^2/cos
again clear fractions:
1-cos^2 = sin^2
and that you should recognize.

Generally, though, one works with one side of the equation until it equals the other side. Assuming no mistakes were made on the way, this shows that the two original sides were equivalent.

It's not always obvious how to do that, though. It's also common to work on both sides at once, trying to simplify things so they come out the same.

(1+cos)/cos = tan^2/(sec-1)
clear fraction on the left, substitute on right
sec + 1 = (sec^2 - 1)/(sec-1)
sec + 1 = (sec+1)(sec-1)/(sec-1)
sec+1 = sec+1

At this point both sides have been reduced to the same expression, showing that their original values must have been equal.