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

1 Answer
Phillip E.
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

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