# 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 $\tan 2 u = \frac{2 t}{1 - {t}^{2}}$, $\sin 2 u = \frac{2 t}{1 + {t}^{2}}$ and $\cos 2 u = \frac{1 - {t}^{2}}{1 + {t}^{2}}$
The left hand side becomes $\left(1 + \frac{2 t}{1 - {t}^{2}}\right) \left(1 - \frac{2 t}{1 + {t}^{2}}\right)$ this becomes $\frac{\left(1 - {t}^{2} + 2 t\right) \left(1 + {t}^{2} - 2 t\right)}{\left(1 - {t}^{2}\right) \left(1 + {t}^{2}\right)}$ which is $\frac{\left(1 - {t}^{2} + 2 t\right) {\left(1 - t\right)}^{2}}{\left(1 - {t}^{2}\right) \left(1 + {t}^{2}\right)}$ which is $\frac{\left(1 - {t}^{2} + 2 t\right)}{\left(1 + {t}^{2}\right)}$ which means that there is no identity