How do you verify the identity #(1 + tan2u)(1 - sin2u) = 1#?

1 Answer
May 31, 2015

There is no identity

Let #t=tan u# then using half angle formulae #tan 2u=(2t)/(1-t^2)#, #sin 2u= (2t)/(1+t^2)# and #cos 2u = (1-t^2)/(1+t^2)#
The left hand side becomes #(1+(2t)/(1-t^2))(1-(2t)/(1+t^2))# this becomes #((1-t^2+2t)(1+t^2-2t))/((1-t^2)(1+t^2))# which is #((1-t^2+2t)(1-t)^2)/((1-t^2)(1+t^2))# which is #((1-t^2+2t))/((1+t^2))# which means that there is no identity