Verify that each equation is an identity.
sin(x + y)/cos...

Question

Verify that each equation is an identity.

sin(x + y)/cos(x - y) = (cot x + cot y)/(1 + cot x cot y)

Accepted Answer

Hello, everyone. We are asked if the given equation is an identity, we are given the sine of X minus Y divided by the cosine of X plus Y. And we want to know does that equal the tangent of X minus the tangent of Y divided by one minus the tangent of X tangent of Y. Our answer choices are a yes B no. For this to be an identity, we have to make the left hand side equal the right hand side on the left hand side. I notice that I have the sine of X minus Y. And I recall that there's an identity that states that, that is equal to the sine of X cosine of Y minus the cosine of X sine of Y. I also have the cosine of X plus Y. And there's an identity for that also. And that one is that the cosine of X multiplied by the cosine of Y minus the sine of X multiplied by the sine of Y will equal the same as the cosine of X plus Y. Now looking at that left hand side, thinking about plugging in these identities, I honestly don't think it'll get me very far. So what I'm gonna do instead is I'm working with the right hand side. So our right hand side is the tangent of X minus the tangent of Y divided by one minus the tangent of X tangent of Y. I'm gonna change all of these tangents into sine and cosine. So I recall that tangent is the same as sine divided by cosine. So I'm gonna rewrite this as the sine of X divided by the cosine of X minus the sine of Y divided by the cosine of Y divided by one minus the sine of X divided by the cosine of X multiplied by the sine of Y divided by the cosine of Y. So now that everything's in terms of sine and cosine, I'm going to make a common denominator for my numerator and that'll be the cosine of X cosine of Y. So to do so I multiply my first fraction, top and bottom by the cosine of Y and my second fraction gets multiplied top and bottom by the cosine of X. And when I do so I get in my numerator, the sine of X cosine of Y divided by the cosine of X cosine of Y minus our second fraction. It will be the cosine of X sine of Y divided by the cosine of X cosine of Y in the denominator. I'm going to write one as one for now. And then I'm going to do the multiplication. So I have one minus and we'll have the sign of XS of Y as our numerator. And the cosine of X cosine of Y is our denominator from there. I'm going to rewrite my numerator fractions as a single fraction. So I will have the sine of X cosine of Y minus the cosine of X sine of Y divided by the cosine of X cosine of Y. I'm gonna treat the number one as if it is the cosine of X cosine of Y divided by the cosine of X cosine of Y. So I have a common denominator in my denominator. So then we still subtract the sin of X sine of Y divided by the cosine of X cosine of Y combining the denominator into a single fraction. My numerator remains the sine of X cosign of Y minus the cosine of X sine of Y divided by the cosine of X cosine of Y. And in my denominator, I now have the cosine of X cosine of Y minus the sine of X sine of Y divided by the cosine of X cosine of Y. All right, almost there. I'm gonna go up to the top of the next column. And when I do so, I recognize the denominators in both of my fractions in this complex fraction are the same. They are both the cosine of X cosine of Y. This means they can be reduced and I will be left with the sine of X cosine of Y minus the cosine of X sine of Y divided by the cosine of X cosine of Y minus the sine of X sine of Y. And now thinking back to those identities, I wrote at the very beginning, my numerator here, the sine of X cosine of Y minus the cosine of X sine of Y is equal to the sine of X minus Y. And in the denominator, the cosine of X cosine of Y minus the sine of X sine of Y is equal to the cosine of X plus Y. And that is our left hand side. So we made the right hand side equal the left hand side. So this is answer choice. A yes, it is an identity. Have a nice day.

Struggling with Trigonometry?

Verify that each equation is an identity.
sin(x + y)/cos(x - y) = (cot x + cot y)/(1 + cot x cot y)

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Struggling with Trigonometry?

Verify that each equation is an identity.sin(x + y)/cos(x - y) = (cot x + cot y)/(1 + cot x cot y)

Watch next

Sum and Difference Identities practice set

Verify that each equation is an identity.
sin(x + y)/cos(x - y) = (cot x + cot y)/(1 + cot x cot y)