What is #(1+cosx)/(1+secx)# for #x=pi/3#?

Question

1195 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-02T07:28:24

#(1+cosx)/(1+secx)=cosx# and for #x=pi/3#, it is #1/2#.

#(1+cosx)/(1+secx)#

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

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

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

= #cosx#

Hence #(1+cosx)/(1+secx)=cosx# for all values of #x#

and for #x=pi/3#

#(1+cosx)/(1+secx)=(1+cos(pi/3))/(1+sec(pi/3))#

= #(1+1/2)/(1+2)=(3/2)/3=3/2xx1/3=1/2#

As #cosx=cos(pi/3)=1/2#

for #x=pi/3#, #(1+cosx)/(1+secx)=cosx#