In Exercises 1–60, verify each identity.cot² t------------ = csc ...

Question

In Exercises 1–60, verify each identity.cot² t------------ = csc t - sin tcsc t

Accepted Answer

Hello, everybody. I have A I today that are going to be looking at this question that states determine whether the given identity is true or false where the identity is open parentheses C and squared A minus one closed parentheses multiplied by cotangents of A is equal to open parentheses. C can A minus cosine a closed parentheses divided by sign of A. Now we have two answer choices A being true and B being false. So for a question like this, I'm going to rewrite our equation. We're going to tackle one side of the equation first and then look at the other side of the equation. And basically, at the end, we're just going to see if both sides match up. So I'm going to start with the left hand side of the equation first. So everything I'm going to be talking about right now is completely referring to the left hand side. So the first thing I wanna do is recall that C can, squared A can be rewritten as one plus tangent squared A. This is thanks to Python I densities. So now I can rewrite our equation as one plus tangent squared A minus one multiplied by cotangent A. Now the one, the positive one and the minus one will cancel out and we'll be left with, or actually before we rewrite this. Let's recall once again that co tangents of A can be rewritten as one divided by a tangent of A. So now we can rewrite and we're left with in one parentheses, we'll have open parentheses, tangents squared A closed parentheses multiplied by one divided by tangents, a closed parentheses. Now the denominator tangent A will cancel out with one of the tangents in the tangent squared A and we're left with tangents of A. So we have on the left hand side, we have tangent of A. So now we look at the right hand side and now everything I talk about now will refer to the right hand side of our equation. So first thing, let's recall that C can A can be rewritten as one divided by a cosine A and now I can rewrite our equation. So I have one divided by cosine A or I have open parentheses, one divided by cosine A minus cosine of a closed parentheses divided by sign of A. Now, the numerator can be rewritten in a way to have to kind of combine both of both parts into one fraction by having a common denominator. And to do that, we'll multiply the cosine of A, by a fraction of cosine of A divided by coin of A. And once we combine that we, we can have one fraction that will be open parentheses, one minus cosine squared A in the numerator divided by cosine of A in the denominator. And then we'll have closed parentheses divided by sign of a still. And now we can recall that's one minus cosine squared A is equal to sine squared A. And this is thanks to a Pythagorean identity. So we can rewrite our equation as sine squared A divided by cosine of A and all that is in parentheses. And we'll have all of that divided by sign of A and we'll have that. The sign of A in the denominator will cancel out with the sine squared A in the numerator of the fraction in the numerator. And we're left with sign of a divided by cosine of A which is equal to tangent of A. So we have tangents of A equals tangents of A which checks out, which means that our identity is true. So I hope that was helpful. And until next time.

In Exercises 1–60, verify each identity.cot² t------------ = csc t - sin tcsc t

Key Concepts

Trigonometric Identities

Reciprocal Functions

Pythagorean Identity

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