Question

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

Answer 1

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