How do you prove #(Sec x/Sin x)-(Sin x/Cos x)=Cot x#?

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Answer 1 · 2018-07-27T14:33:28

See the proof below

We need

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

#secx=1/cosx#

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

Therefore,

#LHS=(secx/sinx)-(sinx/cosx)#

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

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

#=cos^2x/(cosxsinx)#

#=cosx/sinx#

#=cotx#

#=RHS#

#QED#