MHB Which of the following are equal to this identity?

[Umar]

Hello, sorry for the constant questions. But here is a question asking which of these are equal to the identity cot(x)/sin(2x).

I managed to find out that this is equal to the third option of the three, however, apparently this option on its own is not the right answer. I can't seem to get the other two options to be equivalent.

Can someone please help and see if there are other equivalencies?

View attachment 5965

 

[Greg]

Choice D is correct. Use $\cos(2x)=\cos^2(x)-\sin^2(x)$ to see why.

 

[I like Serena]

Hello, sorry for the constant questions. But here is a question asking which of these are equal to the identity cot(x)/sin(2x).

I managed to find out that this is equal to the third option of the three, however, apparently this option on its own is not the right answer. I can't seem to get the other two options to be equivalent.

Can someone please help and see if there are other equivalencies?

Hey Umar! ;)

Let's reduce all of them to contain only $\sin x$ and $\cos x$ and simplify them:
$$\frac{\cot x}{\sin(2x)} = \frac{\cos x}{\sin x\cdot 2\sin x\cos x} = \frac{1}{2\sin^2 x} \\
(i)\quad\frac{1}{1-\cos(2x)} = \frac{1}{1-(1-2\sin^2 x)} = \frac{1}{2\sin^2 x} \\
(ii)\quad\frac 12(1+\tan^2 x) = \frac 12\left(1+\frac{\sin^2 x}{\cos^2 x}\right) = \frac 12\cdot \frac{\cos^2 x+\sin^2 x}{\cos^2 x} = \frac{1}{2\cos^2 x} \\
(iii)\quad\frac{1}{2(1-\cos^2 x)} = \frac 1{2\sin^2 x}
$$
How about the equivalencies now?

 

[MarkFL]

I think what I would do here is treat them all, in turn, as prospective identities to be verified.

i) $$\frac{\cot(x)}{\sin(2x)}=\frac{1}{1-\cos(2x)}$$

$$\cot(x)(1-\cos(2x))=\sin(2x)$$

$$1-(1-2\sin^2(x))=2\sin^2(x)$$

$$2\sin^2(x)=2\sin^2(x)$$

This is an identity.

ii) $$\frac{\cot(x)}{\sin(2x)}=\frac{1}{2}(1+\tan^2(x))$$

$$2\cot(x)=\sin(2x)\sec^2(x)$$

$$2\cot(x)=2\cot(x)$$

This is an identity.

iii) $$\frac{\cot(x)}{\sin(2x)}=\frac{1}{2(1-\cos^2(x))}$$

$$2\cot(x)\sin^2(x)=\sin(2x)$$

$$\sin(2x)=\sin(2x)$$

This is an identity.

So, we find all three are identities. :D

 

[Umar]

Hey Umar! ;)

Let's reduce all of them to contain only $\sin x$ and $\cos x$ and simplify them:
$$\frac{\cot x}{\sin(2x)} = \frac{\cos x}{\sin x\cdot 2\sin x\cos x} = \frac{1}{2\sin^2 x} \\
(i)\quad\frac{1}{1-\cos(2x)} = \frac{1}{1-(1-2\sin^2 x)} = \frac{1}{2\sin^2 x} \\
(ii)\quad\frac 12(1+\tan^2 x) = \frac 12\left(1+\frac{\sin^2 x}{\cos^2 x}\right) = \frac 12\cdot \frac{\cos^2 x+\sin^2 x}{\cos^2 x} = \frac{1}{2\cos^2 x} \\
(iii)\quad\frac{1}{2(1-\cos^2 x)} = \frac 1{2\sin^2 x}
$$
How about the equivalencies now?

Thank you so much, I never really thought about doing it that way, but I kept getting close to that. I see, so only the first and third one are equivalent.