Question #c070f

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Answer 1 · 2016-10-09T16:01:11

Using the double angle formulas and combing the fractions and simplifying the algebra

to prove
#sec(2x)+tan(2x)=(1+tan(x))/(1-tan(x))#

take LHS

#sec(2x)+tan(2x)= 1/cos(2x) +2tan(x)/(1-tan^2(x))#

#=1/(cos^2(x)-sin^2(x))+2tan(x)/(1-tan^2(x))#

Divide every term of the first fraction by #cos^2(x)#

=#(1/cos^2(x))/(cos^2(x)/cos^2(x)-sin^2(x)/cos^2(x))+2tan(x)/(1-tan^2(x))#

#=sec^2(x)/(1-tan^2(x))+2tan(x)/(1-tan^2(x))#

#=(1+tan^2(x))/(1-tan^2(x))+2tan(x)/(1-tan^2(x))#

#=(1+2tan(x)+tan^2(x))/(1-tan^2(x))#

#=(1+tan(x))^2/((1-tan(x))(1+tan(x))#

cancelling a #(1+tan(x)) # bracket

#=(1+tan(x))/(1-tan(x)#

as required.