Beginner's Verifying Trig Identity

[courtbits]

\(\displaystyle (\cot \theta)(\sin \theta)\)

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

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

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

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

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

OR

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

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

The answer choices are ~

a.) \(\displaystyle \tan \theta\)
b.) \(\displaystyle \cos \theta\)

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

 

[kaliprasad]

\(\displaystyle (\cot \theta)(\sin \theta)\)

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

No

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

 

[courtbits]

No

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

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

 

[kaliprasad]

\(\displaystyle (\cot \theta)(\sin \theta)\)

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

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

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

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

\
now you are right till this point then the mistake

 

[courtbits]

\
now you are right till this point then the mistake

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

 

[kaliprasad]

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

yes but 1 in denominator is unnecessary and result becomes

\(\displaystyle \cos \theta\)

 

[courtbits]

yes but 1 in denominator is unnecessary and result becomes

\(\displaystyle \cos \theta\)

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

 

[suluclac]

tan θ = sin θ/cos θ