Prove that #(1+sinx)/(1-sinx)=(secx+tanx)^2#?

3 Answers
Oct 13, 2016

Using the identities

  • #sin^2(x)+cos^2(x) = 1 => 1-sin^2(x) = cos^2(x)#
  • #(a-b)(a+b) = a^2-b^2#

we have

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

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

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

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

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

#=(sec(x)+tan(x))^2#

Oct 13, 2016

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

Explanation:

Let us start from right hand side.

#(secx+tanx)^2#

= #(1/cosx+sinx/cosx)^2#

= #((1+sinx)/cosx)^2#

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

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

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

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

Oct 13, 2016

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

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

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

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

#=((1+sinx)/cosx)^2#

#=(1/cosx+sinx/cosx)^2#

#=(secx+tanx)^2#

Proved