Verify that each equation is an identity.
(tan(α + β) - ...

Question

Verify that each equation is an identity.

(tan(α + β) - tan β)/(1 + tan(α + β) tan β) = tan α

Accepted Answer

Hello, everyone. We are asked if the given equation is an identity, we are given the tangent of X minus Y plus the tangent of Y, all of which is divided by one minus the tangent of X minus Y multiplied by the tangent of Y equals the tangent of X. Our answer choices are a yes B no. For this to be an identity, we need to make the left side match the right side. So the left hand side has most of the information. So I'm gonna start there. The first thing I recall is that I have an identity for the tangent of X minus Y that states that it's the tangent of X minus the tangent of Y divided by one plus the tangent of X multiplied by the tangent of Y. So I'm gonna take this piece that this identity equals to and put it into our equation for the tangent of X minus Y. So I will have the tangent of X minus the tangent of Y divided by one plus the tangent of X tangent of Y plus the tangent of Y as the numerator. And then in the denominator, I have one minus. And I'm gonna put the identity in again. So I'm gonna put in parentheses, the tangent of X minus the tangent of Y divided by one plus the tangent of X tangent of Y closed parentheses. And all of that's gonna be multiplied by the tangent of Y. OK. So from here, I am going to create a common denominator for my numerator. So in the first part of the fraction, the tangent of X minus the tangent of Y will stay put over one plus the tangent X tangent Y. In the second part, I'm multiplying numerator and denominator by the tangent. I'm sorry of by one plus the tangent of X tangent of Y. So in the numerator, they'll give us the tangent of Y multiplied by one plus the tangent of X tangent of why in our denominator we'll have one plus the tangent of X tangent of why I will simplify that in my next step in my denominator, I'm going to rewrite one as one plus the tangent of X tangent of Y divided by one plus the tangent of X tangent of Y minus. I'm distributing the tangent of Y into the numerator of my second fraction. Putting it back in parentheses. And I have the tangent of X tangent of Y minus the tangent squared of Y divided by one plus the tangent of X tangent of Y. All right. So now I want to write this as a single fraction in my numerator. So I'm going to keep my common denominator in the numerator of one plus the tangent of X tangent of Y. Yeah. And then I will have the tangent of X minus the tangent of why. And I'm gonna distribute in my second fraction. The tangent of lie into the parentheses. So plus the tangent of why plus the tangent of X tangent squared of Y in my denominator, I'm distributing my negative into the parentheses as I combine my like terms or as I write it as one fraction. So again, we'll have a common denominator of one plus the tangent of X tangent of Y. In the numerator, I have one plus the tangent of X tangent of Y minus the tangent of X tangent of Y plus tangent squared Y. OK. So there's a lot going on there. Let's see what I can clean up. I noticed that in my complex fraction, both the numerator and the denominator share the same common denominator, which means those can be canceled out. So we'll be down to one single fraction which is great. I'm gonna combine what I can. So in the numerators numerator, the tangent of X does not have anything to be combined with, but the negative tangent of Y and tangent of Y can be canceled. So this would be plus tangent X tangent squared Y in my denominators frac uh numerator, I can cancel out the tangent of X tangent of Y. So I'm left with one plus the tangent squared of Y. And that looks a lot better already. Next, in my numerator, I noticed I could factor out a tangent of X. So I'd have the tangent of X multiplied by the parentheses. One plus the tangent squared of why. And when I divide that by one plus the tangent squared of Y, it looks like it cancels out leaving me with the tangent of X. And if that is the case, then this is answer choice. A yes, it is an identity. Have a nice day.

Verify that each equation is an identity.
(tan(α + β) - tan β)/(1 + tan(α + β) tan β) = tan α

Key Concepts

Trigonometric Identities

Angle Sum Identity for Tangent

Algebraic Manipulation

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