Verify that each equation is an identity.
1/(sec α - tan...

Question

Verify that each equation is an identity.

1/(sec α - tan α) = sec α + tan α

Accepted Answer

Hello. Today we are going to be verifying whether the given equation is an identity or not. In order to verify that this equation is an identity, 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, we 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 of the equation. The left hand side of the equation is given to us as one divided by co sequent of X minus cotangent of X. Now, currently, there's not much that we can do with the given denominator. However, let's go ahead and rewrite the denominator to be in terms of sign and cosign. In order to do this, we'll be using the trigonometric identities for cos seeking of X and cotangent of X. The trigonometric identity for co seeking of X is given to us as one divided by sine of X. And the trigonometric identity for co tangent of X is given to us as cosine of X divided by sine of X. If we were to use these two trigonometric identities, we can rewrite the left hand side of the equation to be one divided by one divided by s of X minus cosine of X divided by sine of X. Since both of these fractions have the same denominator of sin of X, we can combine both of the quantities in the denominator together. And that will allow us to rewrite the denominator as one minus cosine of X divided by sine of X. Now, this is in the form of a complex fraction. In order to simplify the complex fraction, we need to multiply the denominator by its reciprocal. And since we are multiplying the denominator by the reciprocal, we will multiply the numerator by the same reciprocal, the reciprocal of the denominator will be sin of X divided by one minus cosine of X. And we will be multiplying the numerator by the same quantity. This will allow us to reduce the denominator to be just one, leaving us with one multiplied by sine of X divided by one minus cosine of X in the numerator. This will simplify the left hand side to be sign of X divided by one minus cosine of X. Now, currently, there isn't much that we can do with one minus cosine of X in the denominator. What we could do though is we can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of the denominator will be one plus cosine of X and will be multiplying both en numerator and denominator by this conjugate. This will allow us to rewrite the denominator as one minus cosine of X multiplied by one plus cosine of X. That will evaluate to give us one minus cosine squared of X. And in the numerator, if we were to multiply sine of X by one plus cosine of X, we can rewrite the numerator to be sine of X plus sine of X multiplied by cosine of X. Now, in the denominator, we can simplify by using the Pythagorean identity sine squared of X plus cosine squared of X is equal to one. If we were to subtract cosine squared of X to both sides of this identity, we can rewrite the identity to be sine squared of X is equal to one minus cosine squared of X. What this means is that we can simplify the numerator. I mean, I'm sorry, the denominator to just be squad of X. Furthermore, what we can do is we can split apart this fraction because we have a sum in the numerator that will allow us to rewrite the equation or rewrite the left hand side to be sign of X divided by sine squared of X plus sine of X multiplied by cosine of X is divided by sine squared of X. In the left hand fraction. We have one multiple of sine of X present in both the numerator and denominator. That means that we can simplify one signs, one sign of X from both the numerator and denominator. And we can do the same thing to the second fraction that will allow us to rewrite the left hand side of the equation to be one divided by sine of X plus cosine of X divided by sine of X. Now recall earlier, the trigonometric identity for cos seeking of X is equal to one divided by sine of X. And the trigonometric identity for cotangent will be cosine of X divided by sin of X. We have both of those quantities present in the left hand side of the equation. So we can rewrite the left hand side to be co seeking of X plus cotangent of X. But this is not the same as cos seeking of X plus tangent of X. 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 what this means is that 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.

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

Key Concepts

Trigonometric Identities

Reciprocal Functions

Algebraic Manipulation

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