Verify that each equation is an identity.
(1 - cos θ)/(1...

Question

Verify that each equation is an identity.

(1 - cos θ)/(1 + cos θ) = 2 csc² θ - 2 csc θ cot θ - 1

Accepted Answer

Hello, everyone. We are asked if the given equation is an identity, we are given one minus the sine of X divided by one plus the sin of X equals the C can't squared X plus tangent squared X minus two tangent XC can't X. Our answer choices are a yes B no. So for this to be an identity, I need to make the left hand side match the right hand side. In this case, I'm gonna start from the right hand side and see if I can work to make it match the left hand side. So I have C can't squared X plus tangent squared X minus two tangent XC can't X, I recall that I have a Pythagorean identity that states that the tangent squared of X equals the C squared of X minus one. So I'm gonna substitute that in for the tangent squared of X in this expression. So I'll have the C can't squared X plus C can't squared X minus one minus two tangent XC can't X. So let me combine my LA terms and I'll have two C can't squared X minus one minus two tangent XC can't X from there. I think I want to try to get this all in terms of sign and cosign because I think that will be the quickest way to get back to the sign. So rewriting everything in terms of sign and cosign two C squared X would be the same as saying two divided by the cosine squared of X because those are reciprocal identities minus one minus two, which will be multiplied by the sine of X divided by the cosine of X and multiplied by the CX which is one divided by the cosine of X from there. I'm gonna multiply together my fractions that are at the end of this expression. So I'll have two divided by the cosine squared of X minus one minus. Remember that two in front of these parentheses is like two divided by one. So this would be two sine X divided by cosine squared X. I noticed that two of my three terms have a denominator of cosine squared X. So I'm going to rewrite one as cosine squared X divided by cosine squared X. That way, everything has a common denominator. So I'll have two divided by cosine squared X minus cosine squared X divided by cosine squared X minus two sine X divided by cosine squared X. So now that everything has a common denominator, let's subtract. So we will have two minus cosine squared X minus two sine X, all of which is divided by cosine squared X thinking through another Pythagorean identity. I recall that cosine squared X would be equal to one minus sine squared X. So let's see if we can apply that in both our numerator and denominator. So in our numerator, I'd have two minus and then in parentheses, one minus sine squared X closed parentheses minus two sine X. In my denominator, I'd have one minus sine squared X. I'm going to distribute the negative into my parentheses in the numerator. So I have two minus one plus sine squared X minus two sx. So it looks like I have some like terms I can combine there, which I'll do in one second. It's still over one minus sine squared X combining my like terms and my numerator. And I'm gonna rearrange this a little bit. I'll have sine squared X minus two sine X plus one divided by one minus sine squared X. I recognize my numerator is now a perfect square trinomial. I can factor this to be and in parentheses, I'll probably write it a slightly different way, but it'll be OK. It will be one minus sine X squared. And that would be divided by, well, one minus sine squared X is the product of a sum in a different. So I could write this as one plus sine X multiplied by one minus sine X. And when I reduce, I can reduce the one minus Sinex in my denominator with one of the ones in the numerator leaving me with one minus sin X divided by one plus sin X. And that does match my left hand side. So in the end, this is answer choice. A yes, it is an identity. Have a nice day.

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Verify that each equation is an identity.
(1 - cos θ)/(1 + cos θ) = 2 csc² θ - 2 csc θ cot θ - 1