In Exercises 1–60, verify each identity.csc θ - sin θ = cot θ cos...

Question

In Exercises 1–60, verify each identity.csc θ - sin θ = cot θ cos θ

Accepted Answer

Hello everybody. I have night today today, we're going to look at this question that states determine whether the given identity is true or false where the given identity is C can of theta multiplied by co tangent there, you know, is equal to sign of theta plus cotangents of theta multiplied by cosine of the. Now we only have two answer choices A being true and B being false. Now for a question like this where we have to set or we have to prove that both sides of our equation are equal to each other. I want to tackle each side of the equation first. So I'm going to tackle the left hand side first and then we'll talk about the right hand side. So everything that I'm going to be saying now is specifically talking about the left hand side of our equation. So focusing on that, the first thing we'll do is recall and what we call that C cans of theta is equal to one divided by cosine of theta. And we can also recall that co tangent of the is equal to one divided by tangents of theta. And now we can rewrite and we'll be left with open parentheses, one divided by cosine of theta closed parentheses, multiplying an open parentheses, one divided by tangents of theta closed parentheses. Now, if I recall once again, I know that tangent of theta is equal to sign of theta divided by cosine of theta. And once again, I can rewrite and we'll be left with open parentheses, one divided by cosine of theta that stays the same. And now we're multiplying an open parentheses. And if we had one divided by tangent data, and we said that tangent data is sine theta divided by cosine data. One divided by Tanya, theta is the reciprocal of that. So we have cosine of theta divided by sine of theta and we'll close the parentheses. Now, in this multiplication in between the two parentheses, we'll have that cosign of data cancels out and we're left with one divided by sine of theta. And I know that one divided by sine of theta is COCA theta. And that will be it for the left hand side of my equation. So all I have to prove now is that the right hand side of my equation leads to sequence of theta. So we're talking about the right hand side. Now, the first thing we'll do is recall once again. So we'll recall that co tangent of a theta is equal to one divided by tangents of the. So we can rewrite. And if we rewrite, we're left with a sign of the plus open parentheses, one divided by tangents of theta closed parentheses multiplied by cosine of theta. Now remember in the left hand side where we said tangent of data is equal to a sine theta divided by cosine data. So if we have one divided by tangent data, we just take the reciprocal as we did on the left hand side. So we can rewrite once again and we're left with sine of theta plus open parentheses, cosine of theta divided by s sign of theta, closed parentheses, multiplied by open parentheses, cosine of theta. Now, if we multiply both of our parentheses, we, we will now have a sign of the plus cosine squared theta divided by sine of theta. And now to combine, basically, I'm going to combine both of these expressions into one fraction. And to do that, I need to find a common denominator. So I'll multiply the sin of theta by a, by a fraction of sine of the divided by sin of theta. So they have the same denominator. And once we combine those two fractions, we'll have sine squared, theta plus cosine squared, theta and all that divided by a sign of theta. Now we can recall our Python identities and we recall that sine squared theta plus cosine squared, theta is equal to one. So we can rewrite once again. And we're left with one divided by sign of theta, which we said on the left hand side is equal to c cans of theta. So this proves that both sides of our equation are equal to each other, which means that our identity is true or as a choice. A so I hope that was helpful. And until next time.

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In Exercises 1–60, verify each identity.csc θ - sin θ = cot θ cos θ

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