How do you simplify #(1+cos y)/(1+sec y)#?

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Answer 1 · 2015-08-05T22:00:47

#(1+cosy)/(1+secy)=cosy#

#secy=1/cosy#, therefore we have:

#(1+cosy)/(1+secy)=(cosy/cosy)((1+cosy)/(1+1/cosy))=#
#cosy((1+cosy)/(1+cosy))=cosy#

Answer 2 · 2015-08-06T01:23:26

Since #secy = 1/cosy#, you can rewrite this as:

#(1+cosy)/(1+(1/cosy))#

Thus, multiply through by #((cosy)/(cosy))#:

#= (cosy+cos^2y)/(cosy+1)#

#= cosy(cancel((1+cosy)/(cosy+1)))#

#= color(blue)(cosy)#