How do you simplify the expression $\frac{{cot}^{2} x}{csc x - 1}$ $cot^2x/(cscx-1)$ ?

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Answer 1 · 2016-10-15T09:39:24

$= csc θ + 1$ $=csctheta+1$

using the identity

${cot}^{2} θ + 1 = {csc}^{2} θ$ $cot^2theta+1=csc^2theta$

we have:

$\frac{{cot}^{2} θ}{csc - 1} = \frac{{csc}^{2} θ - 1}{csc θ - 1}$ $cot^2theta/(csc-1)=(csc^2theta-1)/(csctheta-1)$

using difference of squares in the numerator

$= \frac{(csc θ + 1) (csc θ - 1)}{csc θ - 1}$ $=((csctheta+1)(csctheta-1))/(csctheta-1)$

$=((csctheta+1)cancel((csctheta-1)))/(cancel((csctheta-1)))$

$=csctheta+1$

How do you simplify the expression cot^2x/(cscx-1)cot2xcscx−1cot^2x/(cscx-1)?