How do you prove #csc (-x) / sec ( - x) =- cot x#?

1 Answer
Mar 17, 2018

We're trying to prove that #csc(-x)/sec(-x)=-cotx#.

You'll need to use reciprocal identities:

#secx=1/cosx#

#cscx=1/sinx#

and also angle difference formulae:

#sin(x-y)=sinxcosy-cosxsiny#

#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:

#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!