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How do you prove #(1+sinx)/(1-sinx)=(secx+tanx)^2#?

2 Answers
Oct 3, 2016

Answer:

see below

Explanation:

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

Right Side #=(secx+tanx)^2#

#=(secx+tanx)(secx+tanx)#

#=sec^2x+2secxtanx+tan^2x#

#=1/cos^2x +2*1/cosx *sinx/cosx +sin^2x/cos^2x#

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

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

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

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

#=# Left Side

Answer:

A simpler derivation is given below :

Explanation:

#{1+sin x}/{1-sin x} = {1+sin x}/{1-sin x} times {1+sin x}/{1+sin x} = {(1+sin x)^2}/{1-sin^2 x} = {(1+sin x)^2}/{cos^2x} = ({1+sin x}/{cos x})^2 = (sec x + tan x)^2#