Question #ca3a5

1 Answer
sjc
Dec 5, 2017

#=1+cotx#

Explanation:

to simplify

#(sin^2x-cos^2x)/(sin^2x-sinx)#

we will factorise the numerator by difference of squares, and teh denominator bya common factor

#=((sinx+cosx)cancel((sinx-cosx)))/(sinxcancel((sinx-cosx)))#

#=(sinx+cosx)/sinx=sinx/sinx+cosx/sinx#

#=1+cotx#

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