NIMCET 2009 Paper

If $\theta ={\tan }^{-1}\frac{1}{1+2}+{\tan }^{-1}\frac{1}{1+(2)(3)}+{\tan }^{-1}\frac{1}{1+(3)(4)}+\cdots +{\tan }^{-1}\frac{1}{1+n(n+1)}$, then $\tan \theta$ is equal to: