How do you prove #Tan^2(x/2+Pi/4)=(1+sinx)/(1-sinx)#?

1 Answer
Apr 23, 2016

Proof below (it's a long one)

Explanation:

Ill work this backwards (but writing doing it forward would work as well):
#(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#
Then substitute in #t# formula (Explanation below)
#=((1+(2t)/(1+t^2))/((1-t^2)/(1+t^2)))^2#
#=(((1+t^2+2t)/(1+t^2))/((1-t^2)/(1+t^2)))^2#
#=((1+t^2+2t)/(1-t^2))^2#
#=((1+2t+t^2)/(1-t^2))^2#
#=((1+t)^2/(1-t^2))^2#
#=((1+t)^2/((1-t)(1+t)))^2#
#=((1+t)/(1-t))^2#
#=((1+tan(x/2))/(1-tan(x/2)))^2#
#=((tan(pi/4)+tan(x/2))/(1-tan(x/2)tan(pi/4)))^2# Note that: (#tan(pi/4)=1)#
#=(tan(x/2+pi/4))^2#
#=tan^2(x/2+pi/4)#

T FORMULAE FOR THIS EQUATION:
#sinx=(2t)/(1-t^2)#, #cosx=(1-t^2)/(1+t^2)#, where #t=tan(x/2)#