How do you verify #(cosX+sinX)/(cscX+secX) = (cosX)(sinX)#?

1 Answer
Feb 22, 2016

Using the definitions of secant and cosecant

  • #csc(x) = 1/sin(x)#
  • #sec(x) = 1/cos(x)#

We have

#(cos(x)+sin(x))/(csc(x)+sec(x)) = (cos(x)+sin(x))/(1/sin(x)+1/cos(x))#

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

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

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

#=cos(x)sin(x)#

(Note that this identity is only true where secant and cosecant are defined, that is, where #sin(x)!=0# and #cos(x)!=0#)