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:

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

$=1/cos theta=sec theta$

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