Question #93fd2

2 Answers
sente
Oct 9, 2016

We will use the following trigonometric identities:

  • #sec(theta) = 1/cos(theta)#
  • #tan(theta) = sin(theta)/cos(theta)#
  • #sin(2theta) = 2sin(theta)cos(theta)#
  • #cos(2theta) = cos^2(theta)-sin^2(theta)#
  • #cos^2(theta)+sin^2(theta) = 1#

as well as the following special products:

  • #a^2+2ab+b^2 = (a+b)^2#
  • #a^2-b^2 = (a+b)(a-b)#

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

#=(1+sin(2x))/cos(2x)#

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

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

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

#=(cos(x)+sin(x))/(cos(x)-sin(x))#

#=((cos(x)+sin(x))/cos(x))/((cos(x)-sin(x))/cos(x))#

#=(1+sin(x)/cos(x))/(1-sin(x)/cos(x))#

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

P dilip_k
Oct 9, 2016

#LHS=sec2x+tan2x#

#=1/cos(2x)+tan2x#

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

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

#=(1+tanx)^2/((1+tanx)(1-tanx))#

#=(1+tanx)/(1-tanx)=RHS#

proved

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