How do you prove $\frac{csc (- x)}{sec (- x)} = - cot x$ $csc (-x) / sec ( - x) =- cot x$ ?

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Answer 1 · 2018-03-17T06:58:30

We're trying to prove that $\frac{csc (- x)}{sec (- x)} = - cot x$ $csc(-x)/sec(-x)=-cotx$ .

You'll need to use reciprocal identities:

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

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

and also angle difference formulae:

$sin (x - y) = sin x cos y - cos x sin y$ $sin(x-y)=sinxcosy-cosxsiny$

$cos (x - y) = cos x cos y + sin x sin y$ $cos(x-y)=cosxcosy+sinxsiny$

Here's the actual problem. I'll be manipulating the right side of the equation until it equals the right:

$L H S = \frac{csc (- x)}{sec (- x)}$ $LHS=csc(-x)/sec(-x)$

$color(white)(LHS)=(quad1/sin(-x)quad)/(1/(cos(-x))$

$color(white)(LHS)=1/sin(-x)*cos(-x)/1$

$color(white)(LHS)=cos(-x)/sin(-x)$

$color(white)(LHS)=cos(0-x)/sin(0-x)$

$color(white)(LHS)=(cos0cosx+sin0sinx)/(sin0cosx-cos0sinx)$

$color(white)(LHS)=(1*cosx+0*sinx)/(0*cosx-1*sinx)$

$color(white)(LHS)=(1*cosx+color(red)cancelcolor(black)(0*sinx))/(color(red)cancelcolor(black)(0*cosx)-1*sinx)$

$color(white)(LHS)=(1*cosx)/(-1*sinx)$

$color(white)(LHS)=cosx/(-1*sinx)$

$color(white)(LHS)=-1*cosx/sinx$

$color(white)(LHS)=-1*cotx$

$color(white)(LHS)=-cotx$

$color(white)(LHS)=RHS$

That's the proof. Hope this helped!

How do you prove csc (-x) / sec ( - x) =- cot xcsc(−x)sec(−x)=−cotxcsc (-x) / sec ( - x) =- cot x?