How do you verify #(1+csc(x))/(cot(x)+cos(x)) = sec(x)#?

1 Answer
May 21, 2016

On the left side,

#(1+cscx)/(cotx+cosx)#

Rewrite #cscx# and #cotx# in terms of #sinx# and #cosx#.

#=(1+color(blue)(1/(sinx)))/(color(purple)((cosx)/(sinx))+cosx)#

Simplify.

#=((sinx+1)/(sinx))/((cosx+sinxcosx)/(sinx))#

#=(sinx+1)/(sinx)*(sinx)/(cosx+sinxcosx)#

#=(sinx+1)/(color(red)cancelcolor(black)(sinx))*(color(red)cancelcolor(black)(sinx))/(cosx+sinxcosx)#

#=(sinx+1)/(cosx+sinxcosx)#

Factor out #cosx# from the denominator.

#=(sinx+1)/(cosx(1+sinx))#

Simplify.

#=color(red)cancelcolor(black)(sinx+1)/(cosxcolor(red)cancelcolor(black)((1+sinx)))#

#=1/cosx#

#=secx#