How do you prove $(\frac{sec x}{sin x}) - (\frac{sin x}{cos x}) = cot x$ $(Sec x/Sin x)-(Sin x/Cos x)=Cot x$ ?

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Answer 1 · 2018-07-27T14:33:28

See the proof below

We need

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

$sec x = \frac{1}{cos x}$ $secx=1/cosx$

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

Therefore,

$L H S = (\frac{sec x}{sin x}) - (\frac{sin x}{cos x})$ $LHS=(secx/sinx)-(sinx/cosx)$

$= (\frac{1}{cos x sin x}) - (\frac{sin x}{cos x})$ $=(1/(cosxsinx))-(sinx/cosx)$

$= \frac{1 - {sin}^{2} x}{cos x sin x}$ $=(1-sin^2x)/(cosxsinx)$

$= \frac{{cos}^{2} x}{cos x sin x}$ $=cos^2x/(cosxsinx)$

$= \frac{cos x}{sin x}$ $=cosx/sinx$

$= cot x$ $=cotx$

$= R H S$ $=RHS$

$Q E D$ $QED$

How do you prove (secxsinx)−(sinxcosx)=cotx(Sec x/Sin x)-(Sin x/Cos x)=Cot x?