Prove the following identities.

Question

$an2 heta=\frac{2 an heta}{1- an^2 heta}$

Accepted Answer

Hey, everyone. In this problem we're asked if the following trigonometric equation is an identity, and the equation given is that cotangent of 2 x is equal to cotangent squared of x minus 1, all divided by 2 cotangent of x. Here we have 2 answer choices, option a, it is an identity, and option b, it is not an identity. So when we have a problem like this and we want to figure out if this is an identity, what the question is really asking is, is this equation true for every x value that it is defined for? If that's the case, then it is an identity. So let's look at the left hand side of our equation and see if we can write it as the right hand side of our equation. Okay, so we have cotangent of 2 x. Let's try to write it in the same form as the right hand side. Well, cotangent of 2 x, recall that cotangent is equal to 1 divided by tangent. So we can write this as 1 divided by tangent of 2x, and we know that tangent of x is equal to co sine of x divided by cosine of x. And that means that one divided by the tangent is going to be cosine of 2x divided by sine of 2x. We're just working through this step by step and we're trying to again write it like the right hand side. On the right hand side in the numerator, we have 2 separate terms being subtracted. So let's use our double angle identities and try to write cosine of 2x and sine of 2x in terms of something else. Now cosine of 2x, recall that we can write as cosine squared of x minus sine squared of x. And in the denominator, we can write sine of 2x as 2 multiplied by cosine of x multiplied by sine of x. All right, so we're getting there. Okay, we now have at least some two terms subtracted in the numerator. In the numerator we have cotangent squared of x minus 1. So we want this minus one term, so if we look at our numerator, if we were to divide the numerator by sine squared of x, that would give us that minus one term, and it actually turns out that it gives us that cotangent squared of x as well, because we have cosine squared of x divided by sine squared of x. So we want to go ahead and divide the numerator by sine squared. If we do that to the numerator, we have to do it to the denominator as well. And so we can write this as cosine squared of x divided by and actually, let's write it out completely. So cosine squared of x minus sine squared of x, all divided by sine squared x. And all of that is divided by 2 cosine x sine of x divided by sine squared of x. Alright. Now let's simplify within the numerator and within the denominator. In the first term, we have cosine squared of x divided by sine squared of x. Then we have minus sine squared x divided by sine squared x, which is just 1. And all of that is going to be divided by 2 cosine x divided by sine of x. Alright, let's simplify again. Now we've already used this information, but we did it backwards. Okay. We have cosine of x divided by sine of x, that's equal to cotangent. Okay. They're squared, and so what we have in the numerator is cotangent squared of x minus 1, and in the denominator we have 2 cotangent of x. And look at that, that's exactly the right hand side. Okay. So the right side of this equation is equal to the left hand side of this equation or any value of x. It doesn't matter what value of x we use, as long as this equation is defined, it is going to satisfy that equation, and therefore this is an identity. Okay? So going back to our answer choices, the correct answer here is option a, this is an identity. Thanks everyone for watching. I hope this video helped. See you in the next one.

Prove the following identities.
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

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Prove the following identities.﻿tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}tan2θ=1−tan2θ2tanθ​﻿

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Prove the following identities.
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$