How do you factor and simplify #secxcscx-csc^2x#?

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Answer 1 · 2016-08-22T02:28:29

Factor out a #cscx#:

#=>cscx(secx - cscx)#

Simplify using the reciprocal identities #csctheta = 1/sintheta# and #sectheta = 1/costheta#.

#=>1/sinx(1/cosx - 1/sinx)#

#=> 1/(sinxcosx) - 1/(sin^2x)#

#=> sinx/(sin^2xcosx) - cosx/(sin^2xcosx)#

#=>(sinx - cosx)/(sin^2xcosx)#

#=>csc^2xsecx(sinx - cosx)#

This is as far as we can go. It should be noted that #x !=2pin, pi/2 + 2pin#

Hopefully this helps!