$\tan \left(\frac{\pi }{4}+\theta \right)\cdot \tan \left(\frac{3\pi }{4}+\theta \right)=$

$\left[\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\cdot \tan \theta }\right]\left[\frac{\tan \frac{3\pi }{4}+\tan \theta }{1-\tan \frac{3\pi }{4}\cdot \tan \theta }\right]$ $\left[\because \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\cdot \tan B}\right]$

$=\frac{1+\tan \theta }{1-\tan \theta }\times \frac{-1+\tan \theta }{1+\tan \theta }$

$=\frac{(1+\tan \theta )}{(1-\tan \theta )}\times \frac{-(1-\tan \theta )}{\left(1+\tan \theta \right)}$

$=-1$