How do you simplify #cottheta/costheta#?

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Answer 1 · 2016-12-19T12:16:57

#cot theta/cos theta = csc theta#

#cot theta/cos theta = 1/tan theta * 1/cos theta#

#= cos theta/sin theta * 1/cos theta#

#= cancel cos theta/sin theta * 1/cancel cos theta = 1/sin theta#

#= csc theta#

Answer 2 · 2016-12-19T12:46:35

#csc(theta)#

We have: #(cot(theta)) / (cos(theta))#

First, let's apply the standard trigonometric identity #cot(theta) = (cos(theta)) / (sin(theta))#:

#= ((cos(theta)) / (sin(theta))) / (cos(theta))#

#= (cos(theta)) / (sin(theta) cos(theta))#

#= (1) / (sin(theta))#

Then, let's apply the identity #csc(theta) = (1) / (sin(theta))#

#= csc(theta)#