How do you prove # (cscx+cotx)(cscx-cotx)=1#?

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Answer 1 · 2018-03-23T08:48:05

See the proof below

We need

#cscx=1/sinx#

#cotx=cosx/sinx#

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

Therefore,

#LHS=(cscx+cotx)(cscx-cotx)#

#=csc^2x-cos^2x#

#=1/sin^2x-cos^2x/sin^2x#

#=(1-cos^2x)/(sin^2x)#

#=sin^2x/sin^2x#

#=1#

#=RHS#

#QED#