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

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Answer 1 · 2016-10-13T10:27:28

Using the identities

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#

Answer 2 · 2016-10-13T10:27:46

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

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)#

Answer 3 · 2016-10-13T10:31:22

#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