MHB Beginner's Verifying Trig Identity

[courtbits]

$$(\cot \theta)(\sin \theta)$$

So far I understand that you can make
$$(\cot a) \implies (\frac{\cos \theta}{\sin \theta})$$

Then it would come to
$$(\frac{\cos \theta}{\sin \theta})(\sin \theta)$$

I'm stuck at when making $$(\sin \theta)$$ into a fraction.

The sine in between the asterisks is what I mean:
$$(\frac{\cos \theta}{\sin \theta}) *(\sin \theta)*$$

I have no idea if the fraction needs to be:
$$(\frac{1}{\sin \theta})$$

OR

$$(\frac{\sin \theta}{1})$$

I know it's silly to ask over, but also how to proceed the problem.

The answer choices are ~

a.) $$\tan \theta$$
b.) $$\cos \theta$$

I would really like to know which answer it is, and the reason behind it.
*Thanks in advance!

 

[kaliprasad]

$$(\cot \theta)(\sin \theta)$$

$$(\cot a) \implies (\frac{\sin \theta}{\cos \theta})$$
!

No

$$\cot \theta = (\frac{\cos \theta}{\sin \theta})$$

 

[courtbits]

No

$$\cot \theta = (\frac{\cos \theta}{\sin \theta})$$

Wow. x.x A typo error. I will fix it. XD

 

[kaliprasad]

$$(\cot \theta)(\sin \theta)$$

So far I understand that you can make
$$(\cot a) \implies (\frac{\cos \theta}{\sin \theta})$$

Then it would come to
$$(\frac{\cos \theta}{\sin \theta})(\sin \theta)$$

I'm stuck at when making $$(\sin \theta)$$ into a fraction.

The sine in between the asterisks is what I mean:
$$(\frac{\cos \theta}{\sin \theta}) *(\sin \theta)*$$
!

\
now you are right till this point then the mistake

 

[courtbits]

\
now you are right till this point then the mistake

Would it be $$(\frac{\cos \theta}{\sin \theta})(\frac{\sin \theta}{1})$$?

 

[kaliprasad]

Would it be $$(\frac{\cos \theta}{\sin \theta})(\frac{\sin \theta}{1})$$?

yes but 1 in denominator is unnecessary and result becomes

$$\cos \theta$$

 

[courtbits]

yes but 1 in denominator is unnecessary and result becomes

$$\cos \theta$$

Oh OH! I got it! Now I understand! XD Thank you very much!

 

[suluclac]

tan θ = sin θ/cos θ