Prove the identity?

Question

Sin^2x / 1- Cosx = Secx + 1 / Secx

Answer 1

RHS = 1/cosx + cosx

= (cosx + cos ^2x)/cos x

LHS = 1-cos^2x/1-cosx

1-cos^2x/1-cosx = (cosx + cos ^2x)/cos x

1-cos^2x = (1-cosx)(cosx + cos ^2x)/cos x

1-cos^2x = (cosx + cos ^2x - cos ^2x - cos ^3x)/cos x

1-cos^2x = 1 - cos ^2x

Answer 2

This is what it looks to me like you are trying to prove:

(sinx)^2/(1 - cosx) = (secx) + (1/secx)

A quick graph of these two functions show that they are not equal. Are you sure you have the correct identity?

If you are trying to prove:

(sinx)^2/(1 - cosx) = (secx + 1)/secx

this is much easier to show.

(sinx)^2/(1 - cosx) =
[1 - (cosx)^2]/(1 - cosx) = {since (sinx)^2 + (cosx)^2 = 1}
(1 - cosx)(1 + cosx)/(1 - cosx) ={factoring the diff of two squares}
1 + cosx = {canceling the common factor out}
(secx/secx) + (1/secx) {since 1/cosx = secx}
(secx + 1)/(secx) {adding the fractions}
QED

I hope my logic was easy to follow.

In the future, be sure to very carefully write your problem. Don't be afraid to use parenthesis in your questions. They seldom hurt and are almost always helpful.

In looking at other answers that people have given, it looks to me like they are having you manipulate both sides of the equation to get a true statement. This is not the best way and, in fact, is completely wrong to do. That is not to say that you can't do it that way, but it is not correct form. You are asked to prove something. In manipulating both sides, you are assuming that it is true to begin with and are going from that assumption. You should ALWAYS only do things to ONE side of the equation and leave the other alone. Just my two cents.

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Sofarsogood:

You got it right. It would be correct if it were prefaced with 'Assuming a=b.' Since you assume it is true, you are finished with one part of the proof. Now you must assume that it is not true and then show that this leads to a contradiction.

Let's use your method to prove that 0<1. By your reasoning I could just say: Assume that 0<1. Done.

Does that prove it?

I am sticking with my original statement. You CAN NOT effectively PROVE that one thing is equal to another by manipulating both sides of an equation. ALL you know is that LHS = LHS or RHS = RHS. You do NOT know LHS = RHS, this is what you are trying to prove.

This site does not lend itself well to discussions so if you want to continue this email me at my username at yahoo.

Answer 3

My secret for trig: always replace everything with sin and cos, it makes these things much easier.

sin^2(x)/(1 - cos(x)) ?= 1/cos(x) + cos(x) = (1 + cos(x))/cos(x)

Common denominators:
sin^2(x)cosx/(cos(x)(1 - cos(x))) ?= (1 + cos(x))(1 - cos(x))/(cos(x)(1 - cos(x)))

Right side of top is (1 + cos(x))(1 - cos(x)) = 1^2 - cos^2(x) = sin^2(x)

so the numerators and denominators are the same, and the equality holds.
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thegreatdilberto, I disagree with your comment about manipulating both sides. The drawback to that method is it does not give you the identity, you need to know that coming in. However, it is a perfectly legitimate proof as long as care is taken to avoid things like dividing by 0 or signs of roots. If the identity is not true then you will end up with a contradiction. This is a common method of proofs. Would you prefer if it were prefaced with something like "assuming a = b, then...".

If you want a constructive proof you can just read it from bottom to top.

Answer 4

Assume you mean
sin² x / (1 - cos x) = (sec x + 1)/ sec x

on the right, divide and get 1 + 1/sec x = 1 + cos x

on the left, multiply top and bottom by the conjugate of (1 - cos x) which is (1 + cos x). you get

sin² x (1 + cos x)/ (1 - cos² x) =
sin²x (1 + cos x) / sin² x =
1 + cos x

and so you have an identity.