How do you prove #(sec x)/(cotx + tanx) = sinx#?

1 Answer
sente
Apr 30, 2016

Using the definitions of #sec(x), cot(x)#, and #tan(x)#, as well as the identity #sin^2(x)+cos^2(x)=1#, for #sin(x)!=0# and #cos(x)!=0#, we have

#sec(x)/(cot(x)+tan(x)) = (1/cos(x))/(cos(x)/sin(x)+sin(x)/cos(x))#

#=(1/cos(x))/(cos(x)/sin(x)+sin(x)/cos(x))*(cos(x)sin(x))/(cos(x)sin(x))#

#=sin(x)/(cos^2(x)+sin^2(x))#

#=sin(x)/1#

#=sin(x)#

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