How do you simplify #(cot(theta))/ (csc(theta) - sin(theta))#?

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Answer 1 · 2016-09-30T16:03:30

#=(costheta/sintheta)/(1/sintheta - sin theta)#

#=(costheta/sintheta)/(1/sintheta - sin^2theta/sintheta)#

#=(costheta/sintheta)/((1 - sin^2theta)/sintheta#

#=(costheta/sintheta)/(cos^2theta/sintheta)#

#=costheta/sintheta xx sintheta/cos^2theta#

#=1/costheta#

#=sectheta#

Hopefully this helps!

Answer 2 · 2016-09-30T16:10:10

#sec theta#

Since #cot theta=cos theta/sin theta and csc theta =1/sin theta#, the expression becomes:

#(cos theta/sin theta)/(1/sintheta-sin theta)#

that's

#(cos theta/sin theta)/((1-sin^2 theta)/sin theta)#;

then, since #1-sin^2 theta=cos^2 theta#, the expression becomes:

#(cos theta/cancel sin theta)/(cos^2 theta/cancel sin theta)#

#=1/cos theta=sec theta#