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,

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

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

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

$=cos^2x/(cosxsinx)$

$=cosx/sinx$

$=cotx$

$=RHS$

$QED$

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