How do you simplify # (sec y - tan y) (sec y + tan y) / sec y#?

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Answer 1 · 2016-06-09T00:12:28

Apply the identity #sectheta = 1/costheta# and #tantheta = sintheta/costheta#.

#1/cosy - siny/cosy((1/cosy + siny/cosy)/(1/cosy))#

#=((1 - siny)(1 + siny)/cosy)/(cancel(cosy) xx 1/cancel(cosy))#

#= (1 - sin^2y)/cosy#

Apply the pythagorean identity #cos^2theta = 1 - sin^2theta#

#= cos^2y/cosy#

#=cosy#

Hopefully this helps!

Answer 2 · 2016-06-09T05:50:37

#cosy#

Given expression
#=((secy-tany)(secy+tany))/secy#

#=(sec^2y-tan^2y)*cosy#
#=1*cosy=cosy#
[Using formula #(sec^2y-tan^2y)=1 and 1/secy=cosy#]