How do you factor the expression and use the fundamental identities to simplify #sin^2xcsc^2x-sin^2x#?

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Answer 1 · 2017-01-17T07:20:33

The answer is #=cos^2x#

We use

#sin^2x+cos^2x=1#

#cscx=1/sinx#

Therefore,

#sin^2xcsc^2x-sin^2x#

#=cancelsin^2x*1/cancel(sin^2x)-sin^2x#

#=1-sin^2x#

#=cos^2x#