Use basic identities to simplify the expression. cot θ sec θ sin θ
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OpenStudy (anonymous):
Do you know your identites foremost?
OpenStudy (anonymous):
Yeah this one uses the pythagorean identity right
OpenStudy (anonymous):
Not exactly, this most likely looks like something that using the reciprocal identities.
\[\Large \cot \theta= \frac {\cos \theta}{\sin \theta}\]
\[\Large \sec \theta= \frac {1}{\cos \theta} \]
\[\Large \csc \theta=\frac{ 1} {\sin \theta} \]
OpenStudy (anonymous):
So you just replace the cot and sec with it.
OpenStudy (anonymous):
I did that I need help solving it because i got the wrong answer
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OpenStudy (anonymous):
Ok well substituting, we get \[\Large \frac {\cos \theta}{\sin \theta} \times \frac {1}{\cos \theta} \times \sin \theta\]
OpenStudy (anonymous):
|dw:1375057814277:dw| the cos and the sin cancel out and you're left with ?
OpenStudy (anonymous):
1 ok thanks that helped could you you help me out with one more ?
OpenStudy (anonymous):
Sure, np
OpenStudy (anonymous):
Thanks its like the one before Use basic identities to simplify the expression.
\[1/ \cot \theta +\sec \sin\]
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OpenStudy (anonymous):
and the cot is squared
OpenStudy (anonymous):
Open a new thread for the question, more help can be awarded
OpenStudy (anonymous):
so close this one
OpenStudy (anonymous):
Yes,
OpenStudy (anonymous):
Oh and tag me @Seadog12
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OpenStudy (anonymous):
how
OpenStudy (anonymous):
press @ then type my name in the comments like how i tagged you ^^