Let g(x)=f(x)+f(1-x) and f^′ ′(x)>0, x ∈(0,1). If g is decreasing in the interval (0, α) and increasing in the interval (α, 1), then tan^-1(2 α)+…

Let $\displaystyle g(x)=f(x)+f(1-x)$ and $\displaystyle f^{\prime \prime}(x)>0, x \in(0,1)$. If $\displaystyle g$ is decreasing in the interval $\displaystyle (0, \alpha)$ and increasing in the interval $\displaystyle (\alpha, 1)$, then $\displaystyle \tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$ is equal to