How do you verify #(1+sinx+cosx)/(1+sinx-cosx)=(1+cosx)/sinx#?

1 Answer
P dilip_k
Apr 30, 2018

#LHS=(1+sinx+cosx)/(1+sinx-cosx)#

#=(sinx(1+sinx+cosx))/(sinx(1+sinx-cosx))#
#=(sinx+sin^2x+sinxcosx)/(sinx(1+sinx-cosx))#

#=(sinx+1-cos^2x+sinxcosx)/(sinx(1+sinx-cosx))#

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

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

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

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