Verify that each equation is an identity.
2 cos² θ - 1 =...

Question

Verify that each equation is an identity.

2 cos² θ - 1 = (1 - tan² θ)/(1 + tan² θ)

Accepted Answer

Hey, everyone here we are asked is the given equation and identity. We are given one minus two sine squared of A is equal to one minus tangent squared of A, all divided by second squared of A. We have two answer, choice options, answer A yes and answer B no. To determine if our equation is an identity, we need to verify that the left hand side is equivalent to the right hand side. So taking a look at our left hand side, we can recall, we are given one minus two sine squared of A. And now we want to manipulate this expression so that eventually it may match the right hand side of the given equation. So the first thing we can do is split up this two sine squared of A. So we have one minus sine squared of A minus sine squared of A. And now, if we notice the first two terms of our current expression, we have one minus sine squared of A. Well, this follows the Pythagorean identity which states that sine squared of A plus cosine squared of A is equivalent to one rearranging this formula. We have cosine squared of A is equal to one minus sine squared of A. So now we see that we can substitute in cosine squared of A into our current expression. And so we have cosine squared of A minus sine squared of A. And now looking at our current expression, we see that it is comprised of cosine and sine terms, but we want it to include tangent and second terms. So we can begin by multiplying sine squared of A by cosine squared of A divided by cosine squared of A. Now to simplify this, we need to recall that tangent of A is equivalent to sine of A divided by cosine of A. So we can see here we do now have a sine squared of A divided by cosine squared of A. So this gives us tangent squared of A and so rewriting, we have cosine squared of A minus tangent squared of A multiplied by cosine squared of A. And now we can transform our cosine terms to second terms by recalling that cosine of A is equivalent to one divided by second of A. So now using this expression into our current expression, we get one divided by C squared of A minus tangent squared of A multiplied by one divided by see squared of A. And now just combining our terms here, since we now have common denominators, we have one minus tangent squared of A, all divided by second squared of A. And with this final expression, we can refer back to our given equation. Looking at the right hand side of the equation, we can notice it directly matches or is equivalent to the left hand side expression that we have rearranged. So we know that the left hand side is in fact equivalent to the right hand side of the equation. And so from this, we can conclude that yes, our given equation is in fact an identity. So with this, we are left with answer choice. A where again, our final answer is yes. Since the left hand side is equal to the right hand side. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Verify that each equation is an identity.
2 cos² θ - 1 = (1 - tan² θ)/(1 + tan² θ)

Key Concepts

Trigonometric Identities

Cosine and Tangent Functions

Algebraic Manipulation

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