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

Question

11912 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-15T09:39:24

#=csctheta+1#

using the identity

#cot^2theta+1=csc^2theta#

we have:

#cot^2theta/(csc-1)=(csc^2theta-1)/(csctheta-1)#

using difference of squares in the numerator

#=((csctheta+1)(csctheta-1))/(csctheta-1)#

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

#=csctheta+1#