If $\operatorname{Tan}^{-1}\left[\frac{1}{1+2}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(2)(3)}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(3)(4)}\right]+\cdots \cdots+\operatorname{Tan}^{-1}\left[\frac{1}{1+n(n+1)}\right]=\operatorname{Tan}^{-1} \theta$, then $\theta=$