How do you prove #tan(2A)/(1+sec(2A))=tan(A)#?

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Answer 1 · 2015-07-06T11:53:39

My proof is done from RHS to LHS.

#tanA=sinA/cosA#

multiply and divide by 2cosA

#tanA=(2sinAcosA)/(2cos^2A)#

#tanA=(Sin2A)/(cos^2A+cos^2A)#

#tanA=(sin2A)/(1-sin^2A+cos^2A)#

#tanA=(sin2A)/(1+cos2A)#

Divide numerator and denominator with cos2A

#tanA=((sin2A)//cos2A)/((1+cos2A)//cos2A#

#tanA=(tan2A)/(1+sec2A)#

Hence proved

Answer 2 · 2015-07-06T16:55:27

Prove trig identity

#(tan 2a)/(1 + (1/cos (2a)) =# #((sin 2a)/(cos 2a)).((cos 2a)/(cos 2a + 1))#

= #(sin 2a)/(cos 2a + 1)# = #(2sin acos a)/(2cos^2 a) =#

#= sin a/cos a = tan a#