How do you simplify the expression #cot^2x/(cscx+1)#?

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Answer 1 · 2016-09-21T19:59:33

#cscx-1#

#cot^2x/(cscx+1)#

One of the Pythagorean identities is #1+cot^2x=csc^2#

Rearranging:
#cot^2x=csc^2x-1#

Substitute #csc^2x-1# into the numerator for #cot^2x#

#(csc^2x-1)/(cscx+1)#

Factor the numerator by difference of squares:

#((cscx+1)(cscx-1))/(cscx+1)#

Cancel the common factor in numerator and denominator:

#cscx-1#