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!
