Verify that each equation is an identity.
[(sec θ - tan ...

Question

Verify that each equation is an identity.

[(sec θ - tan θ)² + 1]/(sec θ csc θ - tan θ csc θ) = 2 tan θ

Accepted Answer

Hello. Today we are going to be verifying that the given equation is an identity. In order to show this, we want to show that the left hand side of the equation will equal to the right hand side of the equation. In order to do this will be algebraically simplifying and using the help of different trigonometric identities. Let's go ahead and take a look at what is given to us on the left hand side, the left hand side gives us the quantity coin of a minus code tangent of A squared plus one. And this will be divided by co seeking of a minus seeking of a minus cotangent of a multiplied by seeking of a. So let's go ahead and start by algebraically simplifying the numerator in the numerator. We are given a squared quantity. Whenever you see a squared quantity, it is normally good to expand the quantity algebraically. If we were to expand coking of a minus cotangent of A squared, we can rewrite the numerator to be coin squared of A minus two. Kosekin of a multiplied by code tangent of a plus co tangent squared of A plus one. Now let's go ahead and take a look at the denominator. The denominator has two terms, but both of the terms have at least one multiple of seeking of a, what this means is that we can factor out a seeking a from the denominator. And this will allow us to rewrite the denominator to be seeking of a multiplied by the quantity coking of a minus co tangent of a. Now, how can we further simplify the numerator? Let's go ahead and group up the last two terms co tangent squared of A plus one. Recall that we have a trigonometric identity that states that co tangent squared of A plus one is equal to cos and squared of A. If we were to use this trigonometric identity, we can simplify the last two terms to be the single term co seek and squared of A. So we can rewrite the numerator to be co seek and squared of A minus two cos of a multiplied by cotangent of A plus co seek and squared of A. Now notice that we have two positive terms of COK and squared of A. What this means is that we can add the two terms together. This will allow us to simplify the numerator to be two co seek and squared of A minus two co seeking of a multiplied by code tangent of A and this will still be divided by the quantity seeking of a multiplied by co seeking of A minus co tangent of a. Now, if we were to continue working with the numerator, notice that both of the terms of the numerator have a two and at least one multiple of co seek. And of a, what this means is that we can factor out a two coin A from the numerator. This will allow us to rewrite the left hand side of the equation to be two cos seeking of a multiplied by the quantity cos seeking of a minus cotangent of A. And this will be divided by seeking of a multiplied by the quantity co seeking of a minus co tangent of A. Since we have the quantity coking of a minus cotangent of A in both the numerator and denominator, we can eliminate both of these quantities. This will allow us to simplify the left hand side to just be two multiplied by cos seeking of a divided by second of a. Now, if we were to go back to the original equation, recall that the right hand side is given to us as two multiplied by s of a. What this means is that we want to rewrite the left hand side to be in terms of sine and cosine. There are two trigonometric identities that we can use to help us with this process. Recall that the trigonometric identity for coin of A is given to us as one divided by sign of A. And the trigonometric identity for seeking of A is given to us as one divided by cosine of a. If we were to use these two trigonometric identities, we can rewrite the left hand side of the equation to be two multiplied by one divided by sine of A, divided by one divided by cosine of A. We will need to simplify this complex fraction. And in order to simplify the complex fraction, we'll need to multiply both the numerator and denominator by the reciprocal of the denominator. The reciprocal of the denominator will just be cosine of A. And again, this is being multiplied to both the numerator and denominator. This will allow us to reduce the denominator to be just one. And that will leave us with two multiplied by the quantity cosine of A divided by s of A. Now recall that the trigonometric identity for cotangent of A is given to us as cotangent of A is equal to cosine of A divided by sine of A. What this means is that the left hand side of the equation will simplify to be two multiplied by cotangent of A, but this is not equal to two multiplied by sine of A. What this means is that the left hand side of the equation does not equal the right hand side of the equation. Therefore, the given equation is not an identity. And with that being said, the answer to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem and I'll go ahead and see you all in the next video.

Struggling with Trigonometry?

Verify that each equation is an identity.
[(sec θ - tan θ)² + 1]/(sec θ csc θ - tan θ csc θ) = 2 tan θ

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Struggling with Trigonometry?

Verify that each equation is an identity.[(sec θ - tan θ)² + 1]/(sec θ csc θ - tan θ csc θ) = 2 tan θ

Watch next

Introduction to Trigonometric Identities practice set

Verify that each equation is an identity.
[(sec θ - tan θ)² + 1]/(sec θ csc θ - tan θ csc θ) = 2 tan θ